[
  {
    "path": ".gitignore",
    "content": "# Prerequisites\n*.d\n\n# Object files\n*.o\n*.ko\n*.obj\n*.elf\n\n# Linker output\n*.ilk\n*.map\n*.exp\n\n# Precompiled Headers\n*.gch\n*.pch\n\n# Libraries\n*.lib\n*.a\n*.la\n*.lo\n\n# Shared objects (inc. Windows DLLs)\n*.dll\n*.so\n*.so.*\n*.dylib\n\n# Executables\n*.exe\n*.out\n*.app\n*.i*86\n*.x86_64\n*.hex\n\n# Debug files\n*.dSYM/\n*.su\n*.idb\n*.pdb\n\n# Kernel Module Compile Results\n*.mod*\n*.cmd\n.tmp_versions/\nmodules.order\nModule.symvers\nMkfile.old\ndkms.conf\n"
  },
  {
    "path": "LICENSE",
    "content": "                    GNU GENERAL PUBLIC LICENSE\n                       Version 3, 29 June 2007\n\n Copyright (C) 2007 Free Software Foundation, Inc. <http://fsf.org/>\n Everyone is permitted to copy and distribute verbatim copies\n of this license document, but changing it is not allowed.\n\n                            Preamble\n\n  The GNU General Public License is a free, copyleft license for\nsoftware and other kinds of works.\n\n  The licenses for most software and other practical works are designed\nto take away your freedom to share and change the works.  By contrast,\nthe GNU General Public License is intended to guarantee your freedom to\nshare and change all versions of a program--to make sure it remains free\nsoftware for all its users.  We, the Free Software Foundation, use the\nGNU General Public License for most of our software; it applies also to\nany other work released this way by its authors.  You can apply it to\nyour programs, too.\n\n  When we speak of free software, we are referring to freedom, not\nprice.  Our General Public Licenses are designed to make sure that you\nhave the freedom to distribute copies of free software (and charge for\nthem if you wish), that you receive source code or can get it if you\nwant it, that you can change the software or use pieces of it in new\nfree programs, and that you know you can do these things.\n\n  To protect your rights, we need to prevent others from denying you\nthese rights or asking you to surrender the rights.  Therefore, you have\ncertain responsibilities if you distribute copies of the software, or if\nyou modify it: responsibilities to respect the freedom of others.\n\n  For example, if you distribute copies of such a program, whether\ngratis or for a fee, you must pass on to the recipients the same\nfreedoms that you received.  You must make sure that they, too, receive\nor can get the source code.  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Therefore, we\nhave designed this version of the GPL to prohibit the practice for those\nproducts.  If such problems arise substantially in other domains, we\nstand ready to extend this provision to those domains in future versions\nof the GPL, as needed to protect the freedom of users.\n\n  Finally, every program is threatened constantly by software patents.\nStates should not allow patents to restrict development and use of\nsoftware on general-purpose computers, but in those that do, we wish to\navoid the special danger that patents applied to a free program could\nmake it effectively proprietary.  To prevent this, the GPL assures that\npatents cannot be used to render the program non-free.\n\n  The precise terms and conditions for copying, distribution and\nmodification follow.\n\n                       TERMS AND CONDITIONS\n\n  0. Definitions.\n\n  \"This License\" refers to version 3 of the GNU General Public License.\n\n  \"Copyright\" also means copyright-like laws that apply to other kinds of\nworks, such as semiconductor masks.\n\n  \"The Program\" refers to any copyrightable work licensed under this\nLicense.  Each licensee is addressed as \"you\".  \"Licensees\" and\n\"recipients\" may be individuals or organizations.\n\n  To \"modify\" a work means to copy from or adapt all or part of the work\nin a fashion requiring copyright permission, other than the making of an\nexact copy.  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Mere interaction with a user through\na computer network, with no transfer of a copy, is not conveying.\n\n  An interactive user interface displays \"Appropriate Legal Notices\"\nto the extent that it includes a convenient and prominently visible\nfeature that (1) displays an appropriate copyright notice, and (2)\ntells the user that there is no warranty for the work (except to the\nextent that warranties are provided), that licensees may convey the\nwork under this License, and how to view a copy of this License.  If\nthe interface presents a list of user commands or options, such as a\nmenu, a prominent item in the list meets this criterion.\n\n  1. 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For example, Corresponding Source\nincludes interface definition files associated with source files for\nthe work, and the source code for shared libraries and dynamically\nlinked subprograms that the work is specifically designed to require,\nsuch as by intimate data communication or control flow between those\nsubprograms and other parts of the work.\n\n  The Corresponding Source need not include anything that users\ncan regenerate automatically from other parts of the Corresponding\nSource.\n\n  The Corresponding Source for a work in source code form is that\nsame work.\n\n  2. Basic Permissions.\n\n  All rights granted under this License are granted for the term of\ncopyright on the Program, and are irrevocable provided the stated\nconditions are met.  This License explicitly affirms your unlimited\npermission to run the unmodified Program.  The output from running a\ncovered work is covered by this License only if the output, given its\ncontent, constitutes a covered work.  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But this requirement does not apply\nif neither you nor any third party retains the ability to install\nmodified object code on the User Product (for example, the work has\nbeen installed in ROM).\n\n  The requirement to provide Installation Information does not include a\nrequirement to continue to provide support service, warranty, or updates\nfor a work that has been modified or installed by the recipient, or for\nthe User Product in which it has been modified or installed.  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If additional permissions\napply only to part of the Program, that part may be used separately\nunder those permissions, but the entire Program remains governed by\nthis License without regard to the additional permissions.\n\n  When you convey a copy of a covered work, you may at your option\nremove any additional permissions from that copy, or from any part of\nit.  (Additional permissions may be written to require their own\nremoval in certain cases when you modify the work.)  You may place\nadditional permissions on material, added by you to a covered work,\nfor which you have or can give appropriate copyright permission.\n\n  Notwithstanding any other provision of this License, for material you\nadd to a covered work, you may (if authorized by the copyright holders of\nthat material) supplement the terms of this License with terms:\n\n    a) Disclaiming warranty or limiting liability differently from the\n    terms of sections 15 and 16 of this License; or\n\n    b) Requiring preservation of specified reasonable legal notices or\n    author attributions in that material or in the Appropriate Legal\n    Notices displayed by works containing it; or\n\n    c) Prohibiting misrepresentation of the origin of that material, or\n    requiring that modified versions of such material be marked in\n    reasonable ways as different from the original version; or\n\n    d) Limiting the use for publicity purposes of names of licensors or\n    authors of the material; or\n\n    e) Declining to grant rights under trademark law for use of some\n    trade names, trademarks, or service marks; or\n\n    f) Requiring indemnification of licensors and authors of that\n    material by anyone who conveys the material (or modified versions of\n    it) with contractual assumptions of liability to the recipient, for\n    any liability that these contractual assumptions directly impose on\n    those licensors and authors.\n\n  All other non-permissive additional terms are considered \"further\nrestrictions\" within the meaning of section 10.  If the Program as you\nreceived it, or any part of it, contains a notice stating that it is\ngoverned by this License along with a term that is a further\nrestriction, you may remove that term.  If a license document contains\na further restriction but permits relicensing or conveying under this\nLicense, you may add to a covered work material governed by the terms\nof that license document, provided that the further restriction does\nnot survive such relicensing or conveying.\n\n  If you add terms to a covered work in accord with this section, you\nmust place, in the relevant source files, a statement of the\nadditional terms that apply to those files, or a notice indicating\nwhere to find the applicable terms.\n\n  Additional terms, permissive or non-permissive, may be stated in the\nform of a separately written license, or stated as exceptions;\nthe above requirements apply either way.\n\n  8. Termination.\n\n  You may not propagate or modify a covered work except as expressly\nprovided under this License.  Any attempt otherwise to propagate or\nmodify it is void, and will automatically terminate your rights under\nthis License (including any patent licenses granted under the third\nparagraph of section 11).\n\n  However, if you cease all violation of this License, then your\nlicense from a particular copyright holder is reinstated (a)\nprovisionally, unless and until the copyright holder explicitly and\nfinally terminates your license, and (b) permanently, if the copyright\nholder fails to notify you of the violation by some reasonable means\nprior to 60 days after the cessation.\n\n  Moreover, your license from a particular copyright holder is\nreinstated permanently if the copyright holder notifies you of the\nviolation by some reasonable means, this is the first time you have\nreceived notice of violation of this License (for any work) from that\ncopyright holder, and you cure the violation prior to 30 days after\nyour receipt of the notice.\n\n  Termination of your rights under this section does not terminate the\nlicenses of parties who have received copies or rights from you under\nthis License.  If your rights have been terminated and not permanently\nreinstated, you do not qualify to receive new licenses for the same\nmaterial under section 10.\n\n  9. Acceptance Not Required for Having Copies.\n\n  You are not required to accept this License in order to receive or\nrun a copy of the Program.  Ancillary propagation of a covered work\noccurring solely as a consequence of using peer-to-peer transmission\nto receive a copy likewise does not require acceptance.  However,\nnothing other than this License grants you permission to propagate or\nmodify any covered work.  These actions infringe copyright if you do\nnot accept this License.  Therefore, by modifying or propagating a\ncovered work, you indicate your acceptance of this License to do so.\n\n  10. Automatic Licensing of Downstream Recipients.\n\n  Each time you convey a covered work, the recipient automatically\nreceives a license from the original licensors, to run, modify and\npropagate that work, subject to this License.  You are not responsible\nfor enforcing compliance by third parties with this License.\n\n  An \"entity transaction\" is a transaction transferring control of an\norganization, or substantially all assets of one, or subdividing an\norganization, or merging organizations.  If propagation of a covered\nwork results from an entity transaction, each party to that\ntransaction who receives a copy of the work also receives whatever\nlicenses to the work the party's predecessor in interest had or could\ngive under the previous paragraph, plus a right to possession of the\nCorresponding Source of the work from the predecessor in interest, if\nthe predecessor has it or can get it with reasonable efforts.\n\n  You may not impose any further restrictions on the exercise of the\nrights granted or affirmed under this License.  For example, you may\nnot impose a license fee, royalty, or other charge for exercise of\nrights granted under this License, and you may not initiate litigation\n(including a cross-claim or counterclaim in a lawsuit) alleging that\nany patent claim is infringed by making, using, selling, offering for\nsale, or importing the Program or any portion of it.\n\n  11. Patents.\n\n  A \"contributor\" is a copyright holder who authorizes use under this\nLicense of the Program or a work on which the Program is based.  The\nwork thus licensed is called the contributor's \"contributor version\".\n\n  A contributor's \"essential patent claims\" are all patent claims\nowned or controlled by the contributor, whether already acquired or\nhereafter acquired, that would be infringed by some manner, permitted\nby this License, of making, using, or selling its contributor version,\nbut do not include claims that would be infringed only as a\nconsequence of further modification of the contributor version.  For\npurposes of this definition, \"control\" includes the right to grant\npatent sublicenses in a manner consistent with the requirements of\nthis License.\n\n  Each contributor grants you a non-exclusive, worldwide, royalty-free\npatent license under the contributor's essential patent claims, to\nmake, use, sell, offer for sale, import and otherwise run, modify and\npropagate the contents of its contributor version.\n\n  In the following three paragraphs, a \"patent license\" is any express\nagreement or commitment, however denominated, not to enforce a patent\n(such as an express permission to practice a patent or covenant not to\nsue for patent infringement).  To \"grant\" such a patent license to a\nparty means to make such an agreement or commitment not to enforce a\npatent against the party.\n\n  If you convey a covered work, knowingly relying on a patent license,\nand the Corresponding Source of the work is not available for anyone\nto copy, free of charge and under the terms of this License, through a\npublicly available network server or other readily accessible means,\nthen you must either (1) cause the Corresponding Source to be so\navailable, or (2) arrange to deprive yourself of the benefit of the\npatent license for this particular work, or (3) arrange, in a manner\nconsistent with the requirements of this License, to extend the patent\nlicense to downstream recipients.  \"Knowingly relying\" means you have\nactual knowledge that, but for the patent license, your conveying the\ncovered work in a country, or your recipient's use of the covered work\nin a country, would infringe one or more identifiable patents in that\ncountry that you have reason to believe are valid.\n\n  If, pursuant to or in connection with a single transaction or\narrangement, you convey, or propagate by procuring conveyance of, a\ncovered work, and grant a patent license to some of the parties\nreceiving the covered work authorizing them to use, propagate, modify\nor convey a specific copy of the covered work, then the patent license\nyou grant is automatically extended to all recipients of the covered\nwork and works based on it.\n\n  A patent license is \"discriminatory\" if it does not include within\nthe scope of its coverage, prohibits the exercise of, or is\nconditioned on the non-exercise of one or more of the rights that are\nspecifically granted under this License.  You may not convey a covered\nwork if you are a party to an arrangement with a third party that is\nin the business of distributing software, under which you make payment\nto the third party based on the extent of your activity of conveying\nthe work, and under which the third party grants, to any of the\nparties who would receive the covered work from you, a discriminatory\npatent license (a) in connection with copies of the covered work\nconveyed by you (or copies made from those copies), or (b) primarily\nfor and in connection with specific products or compilations that\ncontain the covered work, unless you entered into that arrangement,\nor that patent license was granted, prior to 28 March 2007.\n\n  Nothing in this License shall be construed as excluding or limiting\nany implied license or other defenses to infringement that may\notherwise be available to you under applicable patent law.\n\n  12. No Surrender of Others' Freedom.\n\n  If conditions are imposed on you (whether by court order, agreement or\notherwise) that contradict the conditions of this License, they do not\nexcuse you from the conditions of this License.  If you cannot convey a\ncovered work so as to satisfy simultaneously your obligations under this\nLicense and any other pertinent obligations, then as a consequence you may\nnot convey it at all.  For example, if you agree to terms that obligate you\nto collect a royalty for further conveying from those to whom you convey\nthe Program, the only way you could satisfy both those terms and this\nLicense would be to refrain entirely from conveying the Program.\n\n  13. Use with the GNU Affero General Public License.\n\n  Notwithstanding any other provision of this License, you have\npermission to link or combine any covered work with a work licensed\nunder version 3 of the GNU Affero General Public License into a single\ncombined work, and to convey the resulting work.  The terms of this\nLicense will continue to apply to the part which is the covered work,\nbut the special requirements of the GNU Affero General Public License,\nsection 13, concerning interaction through a network will apply to the\ncombination as such.\n\n  14. Revised Versions of this License.\n\n  The Free Software Foundation may publish revised and/or new versions of\nthe GNU General Public License from time to time.  Such new versions will\nbe similar in spirit to the present version, but may differ in detail to\naddress new problems or concerns.\n\n  Each version is given a distinguishing version number.  If the\nProgram specifies that a certain numbered version of the GNU General\nPublic License \"or any later version\" applies to it, you have the\noption of following the terms and conditions either of that numbered\nversion or of any later version published by the Free Software\nFoundation.  If the Program does not specify a version number of the\nGNU General Public License, you may choose any version ever published\nby the Free Software Foundation.\n\n  If the Program specifies that a proxy can decide which future\nversions of the GNU General Public License can be used, that proxy's\npublic statement of acceptance of a version permanently authorizes you\nto choose that version for the Program.\n\n  Later license versions may give you additional or different\npermissions.  However, no additional obligations are imposed on any\nauthor or copyright holder as a result of your choosing to follow a\nlater version.\n\n  15. Disclaimer of Warranty.\n\n  THERE IS NO WARRANTY FOR THE PROGRAM, TO THE EXTENT PERMITTED BY\nAPPLICABLE LAW.  EXCEPT WHEN OTHERWISE STATED IN WRITING THE COPYRIGHT\nHOLDERS AND/OR OTHER PARTIES PROVIDE THE PROGRAM \"AS IS\" WITHOUT WARRANTY\nOF ANY KIND, EITHER EXPRESSED OR IMPLIED, INCLUDING, BUT NOT LIMITED TO,\nTHE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR\nPURPOSE.  THE ENTIRE RISK AS TO THE QUALITY AND PERFORMANCE OF THE PROGRAM\nIS WITH YOU.  SHOULD THE PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF\nALL NECESSARY SERVICING, REPAIR OR CORRECTION.\n\n  16. Limitation of Liability.\n\n  IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING\nWILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MODIFIES AND/OR CONVEYS\nTHE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES, INCLUDING ANY\nGENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING OUT OF THE\nUSE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED TO LOSS OF\nDATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY YOU OR THIRD\nPARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER PROGRAMS),\nEVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE POSSIBILITY OF\nSUCH DAMAGES.\n\n  17. Interpretation of Sections 15 and 16.\n\n  If the disclaimer of warranty and limitation of liability provided\nabove cannot be given local legal effect according to their terms,\nreviewing courts shall apply local law that most closely approximates\nan absolute waiver of all civil liability in connection with the\nProgram, unless a warranty or assumption of liability accompanies a\ncopy of the Program in return for a fee.\n\n                     END OF TERMS AND CONDITIONS\n\n            How to Apply These Terms to Your New Programs\n\n  If you develop a new program, and you want it to be of the greatest\npossible use to the public, the best way to achieve this is to make it\nfree software which everyone can redistribute and change under these terms.\n\n  To do so, attach the following notices to the program.  It is safest\nto attach them to the start of each source file to most effectively\nstate the exclusion of warranty; and each file should have at least\nthe \"copyright\" line and a pointer to where the full notice is found.\n\n    {one line to give the program's name and a brief idea of what it does.}\n    Copyright (C) {year}  {name of author}\n\n    This program is free software: you can redistribute it and/or modify\n    it under the terms of the GNU General Public License as published by\n    the Free Software Foundation, either version 3 of the License, or\n    (at your option) any later version.\n\n    This program is distributed in the hope that it will be useful,\n    but WITHOUT ANY WARRANTY; without even the implied warranty of\n    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the\n    GNU General Public License for more details.\n\n    You should have received a copy of the GNU General Public License\n    along with this program.  If not, see <http://www.gnu.org/licenses/>.\n\nAlso add information on how to contact you by electronic and paper mail.\n\n  If the program does terminal interaction, make it output a short\nnotice like this when it starts in an interactive mode:\n\n    {project}  Copyright (C) {year}  {fullname}\n    This program comes with ABSOLUTELY NO WARRANTY; for details type `show w'.\n    This is free software, and you are welcome to redistribute it\n    under certain conditions; type `show c' for details.\n\nThe hypothetical commands `show w' and `show c' should show the appropriate\nparts of the General Public License.  Of course, your program's commands\nmight be different; for a GUI interface, you would use an \"about box\".\n\n  You should also get your employer (if you work as a programmer) or school,\nif any, to sign a \"copyright disclaimer\" for the program, if necessary.\nFor more information on this, and how to apply and follow the GNU GPL, see\n<http://www.gnu.org/licenses/>.\n\n  The GNU General Public License does not permit incorporating your program\ninto proprietary programs.  If your program is a subroutine library, you\nmay consider it more useful to permit linking proprietary applications with\nthe library.  If this is what you want to do, use the GNU Lesser General\nPublic License instead of this License.  But first, please read\n<http://www.gnu.org/philosophy/why-not-lgpl.html>.\n"
  },
  {
    "path": "README.md",
    "content": "# Zoho-Interview-questions\nprevious zoho interview questions\n"
  },
  {
    "path": "Set 01 Qns.txt",
    "content": "SET 1\n\nSecond Round: (Normal Programming round)\n1. Print the word with odd letters as\n\nP         M\n R      A\n   O  R\n     G\n  O    R\n R       A\nP          M \n\n\n2. Given a set of numbers like <10, 36, 54,89,12> we want to find sum of weights based on the following conditions\n    1. 5 if a perfect square\n    2. 4 if multiple of 4 and divisible by 6\n    3. 3 if even number\n\nAnd sort the numbers based on the weight and print it as follows\n\n<10,its_weight>,<36,its weight><89,its weight>\n\nShould display the numbers based on increasing order.\n\n\n3. Save the string “WELCOMETOZOHOCORPORATION” in a two dimensional array and search for substring like “too” in the two dimensional string both from left to right and from top to bottom.\n\nw\te\tL\tC\tO\nM\tE\tT\tO\tZ\nO\tH\tO\tC\tO\nR\tP\tO\tR\tA\nT\tI\tO\tn\t  \n\nAnd print the start and ending index as\n\nStart index : <1,2>\n\nEnd index: <3, 2>\n\n\n4. Given a 9×9 sudoku we have to evaluate it for its correctness. We have to check both the sub matrix correctness and the whole sudoku correctness.\n\n\n5. Given a two dimensional array of string like\n\n  <”luke”, “shaw”>\n  <”wayne”, “rooney”>\n  <”rooney”, “ronaldo”>\n  <”shaw”, “rooney”> \n\nWhere the first string is “child”, second string is “Father”. And given “ronaldo” we have to find his no of grandchildren Here “ronaldo” has 2 grandchildren. So our output should be 2.\n\n\nThird Round: (Advanced Programming Round)\n\n\nHere they asked us to create a “Railway reservation system” and gave us 4 modules. The modules were:\n    1. Booking\n    2. Availability checking\n    3. Cancellation\n    4. Prepare chart\nWe were asked to create the modules for representing each data first and to continue with the implementation phase.\n"
  },
  {
    "path": "Set 01-01.print the following pattern.c",
    "content": "  /*  To print the following pattern\n        \tP     M\n                 R   A\n                  O R\n                   G\n                  O R\n                 R   A\n                P     M                   */\n\n#include<stdio.h>\nint main(void)\n{\n\n\tchar str[100],i,j,k,n;\n\tprintf(\"Enter the string : \");\n\tscanf(\"%s\",str);\n\tfor(n=-1;str[n];n++);\n        if(n%2==0)\n        {\n          printf(\"\\n\\nString length is even\\n\\n\");\n          return;\n        }\n\tn=n/2+1;\n\tfor(i=1;i<=n;i++)\n\t{\n\t\tk=0;\n\t\tfor(j=1;j<i;j++)\n\t\t\tprintf(\" \");\n\t\tfor(j=1;j<=2*(n-i)+1;j++)\n\t\t{  if(j==1 || j==2*(n-i)+1)\n\t\t\tprintf(\"%c\",str[i-1+k]);\n\t\t\telse\n\t\t\t\tprintf(\" \");\n\t\t\tk++;\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\n\tfor(i=2;i<=n;i++)\n\t{\n\t\tk=0;\n\t\tfor(j=1;j<=n-i;j++)\n\t\t\tprintf(\" \");\n\t\tfor(j=1;j<=2*i-1;j++)\n\t\t{\n\t\t\tif(j==1 || j==2*i-1) \n\t\t\t\tprintf(\"%c\",str[n-i+k]);\n\t\t\telse\n\t\t\t\tprintf(\" \");\n\t\t\tk++;\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n}\n\n"
  },
  {
    "path": "Set 01-02.find sum of weights based on the following conditions.c",
    "content": "/* Given a set of numbers like <10,36,54,89,12> we want to find sum of weights based on the following conditions\n    1. 5 if a perfect square\n    2. 4 if multiple of 4 and divisible by 6\n    3. 3 if even number\n\nAnd sort the numbers based on the weight and print it as follows\n\n<10,its_weight>,<36,its weight><89,its weight>\n\nShould display the numbers based on increasing order*/\n#include<stdlib.h>\n#include<stdio.h>\nint check(int);\nint main(void)\n{\n\tint *a,i,j,n;\n\tprintf(\"Array size : \");\n        scanf(\"%d\",&n);\n        a=malloc(sizeof(int)*n);\n           printf(\"Enter array elements : \");\n        for(i=0;i<n;i++)\n            scanf(\"%d\",&a[i]);\n\tprintf(\"Array elements : \");\n        for(i=0;i<n;i++)\n            printf(\"%d \",a[i]);\n\tfor(i=0;i<n-1;i++)\n\t\tfor(j=0;j<n-i-1;j++)\n\t\t\tif(check(a[j])>check(a[j+1]))\n\t\t\t\ta[j]=a[j]+a[j+1]-(a[j+1]=a[j]);\n\tprintf(\"The elements after sorting with its weight is : \\n\");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"< %d , %d >\\n\",a[i],check(a[i]));\n        free(a);\n\treturn 0;\n}\n\nint check(int n)\n{\n\tint sum,weight=0,i;\n\tfor(sum=0,i=1;i<n;i+=2)\n\t{\n\t\tsum=sum+i;\n\t\tif(sum==n)\n\t\t\tweight+=5;\n\t}\n\tif(n%4==0 && n%6==0)\n\t\tweight+=4;\n\tif(n%2==0)\n\t\tweight+=3;\n\treturn weight;\n}\n"
  },
  {
    "path": "Set 01-03.Save the string in a two dimensional array and search for substring.c",
    "content": "/* Save the string “WELCOMETOZOHOCORPORATION” in a two dimensional array and search for substring like “too” in the two dimensional string both from left to right and from top to bottom.\n\nw\te\tL\tC\tO\nM\tE\tT\tO\tZ\nO\tH\tO\tC\tO\nR\tP\tO\tR\tA\nT\tI\tO\tn\t  \n\nAnd print the start and ending index as\n\nStart index : <1,2>\n\nEnd index: <3, 2>*/\n#include<stdio.h>\nint sqt(float);\nint main(void)\n{\n // a is 2d array to store the sting\nchar str[100],str1[100],a[100][100];\n //main variables inx =index of string  n= length of string rc=number of rows and coloumn(square matrix)\nint n,rc,i,j,inx,k,c;\n //enter main string to str\nprintf(\"Enter the string : \");\nscanf(\"%s\",str);\n//find length of string\nfor(n=0;str[n];n++);\n // find row and coloumn size by taking square root of n(length of string\nif(n=sqt(n)*sqt(n))\n rc=sqt(n)+1;\nelse\n rc=sqt(n);\n //enter each character in to 2 diamensional array\nfor(inx=0,i=0;i<rc ;i++)\n  for(j=0;j<rc && str[inx];j++)\n   { \n      a[i][j]=str[inx++];\n   }\n// print contents of 2d array\nfor(i=0;i<rc;i++)\n{\n  for(j=0;j<rc;j++)\n   printf(\"%c \",a[i][j]);\n  printf(\"\\n\");\n}\n//Enter the substring as str1\nprintf(\"Enter word to found  :  \");\nscanf(\"%s\",str1);\n//find substring in rows\nfor(i=0;i<rc;i++)\n  for(j=0;j<rc;j++)\n    if(a[i][j]==str1[0])\n    {  for(k=j,c=0;a[i][k]==str1[c] && a[i][k]!='\\0';k++,c++);\n       if(str1[c]=='\\0'){\n          printf(\"\\nStart index : <%d ,%d>   End index: <%d, %d>\\n\",i,j,i,k-1);}\n    }\n//find substring in coloumns\nfor(i=0;i<rc;i++)\n  for(j=0;j<rc;j++)\n    if(a[j][i]==str1[0])\n    { \n      for(k=j,c=0;a[k][i]==str1[c] && a[k][i]!='\\0';k++,c++);\n       if(str1[c]=='\\0'){\n         printf(\"\\nStart index : <%d ,%d>   End index: <%d, %d>\\n\",j,i,k-1,i);}\n    }          \nreturn 0;\n\n}\n\n//user defined function to find square root by Babylonian method\nint sqt(float n)\n{\nfloat x=n,y=1;\nwhile(x-y>0.00001)\n{\n  x=(x+y)/2;\n  y=n/x;\n}\nreturn x;\n}\n"
  },
  {
    "path": "Set 01-05.find no of grandchildren.c",
    "content": "/*  Given a two dimensional array of string like\n\n  <”luke”, “shaw”>\n  <”wayne”, “rooney”>\n  <”rooney”, “ronaldo”>\n  <”shaw”, “rooney”> \n\nWhere the first string is “child”, second string is “Father”. \nAnd given “ronaldo” we have to find his no of grandchildren Here “ronaldo” has 2 grandchildren.\n So our output should be 2. */\n#include<stdlib.h>\n#include<stdio.h>\nint main(void)\n{\nchar *str,***a;\nstr=malloc(50);\nint n,i,j,c=0;\n//enter th number of pair of strings\nprintf(\"Enter the number of pair : \");\nscanf(\"%d\",&n);\n//create 3d array dynamically using malloc\na=malloc(sizeof(char*)*n);     //allocate n character pointer spaces and store addres in a\nfor(i=0;i<n;i++)                 \n{ a[i]=malloc(sizeof(char*)*2); // store 2 chara pointer spaces to every previously allocated n spaces \n  for(j=0;j<2;j++)\n     a[i][j]=malloc(sizeof(char)*50); // allocate 50 bytes to previously allocated 2 char pointer spaces\n}\n//enter the list of strings\nprintf(\"Enter the strings : \");\nfor(i=0;i<n;i++)\n{\n  for(j=0;j<2;j++)\n   scanf(\" %s\",a[i][j]);\n}\n//print length of strings\nprintf(\"The Array is : \\n\");\nfor(i=0;i<n;i++)\n{\n  printf(\"<\");\n  for(j=0;j<2;j++)\n   printf(\"%-10s \",a[i][j]);\n  printf(\">\\n\");\n}\n//enter the grangfather name \nprintf(\"Enter the string : \");\nscanf(\"%s\",str);\n\nprintf(\"Grandchildren of %s :  \",str);\nfor(i=0;i<n;i++)\n  if(strcmp(a[i][1],str)==0)             //Locate grandfather in the list\n      for(j=0;j<n;j++)                   // Find son of the grandfather\n         if(strcmp(a[j][1],a[i][0])==0)  //find grandchildren\n          {  printf(\"%s \",a[j][0]); \n             c++;                        //increment count for every grandchildren   \n          }\nprintf(\"\\nNumber of grandchildren  :  %d\\n\",c);\nfree(a);\nfree(str);\nreturn 0;\n}         \n\n"
  },
  {
    "path": "Set 02-01.Alternate sorting.c",
    "content": "/*  Alternate sorting: Given an array of integers, rearrange the array in such a way that the first element is first maximum and second element is first minimum.\n\n    Eg.) Input  : {1, 2, 3, 4, 5, 6, 7}\n         Output : {7, 1, 6, 2, 5, 3, 4} */\n#include<stdio.h>\nint main(void)\n{\n\tint a[100],i,j,n,t;\n\tprintf(\"Array size : \");\n\tscanf(\"%d\",&n);\n\tprintf(\"Enter array elements : \");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\"%d\",&a[i]);\n\tprintf(\"Array elements : \\n\");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"%d \",a[i]);       \n\tprintf(\"\\n\");\n\t/* Selection sort is used because the 1st element would be fixed after one iteration of outer loop*/ \n\tfor(i=0;i<n-1;i++)                     \n\t\tfor(j=i+1;j<n;j++)\n\t\t{\n\t\t\tif(i%2)                   // if i value is odd then smallest value in remaining index values is goes to ith index\n\t\t\t{  if(a[i]>a[j])\n\t\t\t\ta[i]=a[i]+a[j]-(a[j]=a[i]);\n\t\t\t}\n\t\t\telse                   //  if i value is even then largest value in remaining index values is goes to ith index\n\t\t\t{\n\t\t\t\tif(a[i]<a[j])   \n\t\t\t\t\ta[i]=a[i]+a[j]-(a[j]=a[i]);\n\t\t\t}\n\t\t} \n\tprintf(\"Array elements : \\n\");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"%d \",a[i]);       \n\tprintf(\"\\n\");\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 02-01.Alternate sorting.py",
    "content": "#!/usr/bin/python\n''' Alternate sorting: Given an array of integers, rearrange the array in such a way that the first element is first maximum and second element is first minimum.\n    Eg.) Input  : {1, 2, 3, 4, 5, 6, 7}\n         Output : {7, 1, 6, 2, 5, 3, 4}'''\narr=[] \ndef main():\n     arr_size=input(\"array size : \")\n     print \"elements : \"\n     for i in range(arr_size):\n         arr.append(input())\n     for i in range(len(arr)-1):\n         if i%2==0:\n             for j in range(i,len(arr)):\n                 if arr[i]<arr[j]:\n                     arr[i],arr[j]=arr[j],arr[i]\n         else:\n             for j in range(i,len(arr)):\n                 if arr[i]>arr[j]:\n                     arr[i],arr[j]=arr[j],arr[i]\n     print arr  \n     \nif __name__==\"__main__\":\n        main()\n"
  },
  {
    "path": "Set 02-02.Remove unbalanced parentheses in a given expression.c",
    "content": "/* Remove unbalanced parentheses in a given expression.\n\n    Eg.) Input  : ((abc)((de))\n         Output : ((abc)(de)) \n \n         Input  : (a(b)))(cd)  \n         Output : (a(b))(cd)\n\n\t Input  : (a(b)))(c(d)\n         Output : (a(b))(cd)\n\n         Input  : (ab))(c(d))))\n         Output : (ab)(c(d))\n\n         Input  : (((ab)\n         Output : (ab) \n\n*/\n#include<stdio.h>\nint main(void)\n{\n\tchar str[100];\n\tint i,j,end=0;\n\tprintf(\"enter the string : \");\n\tscanf(\"%s\",str);\n\tfor(i=0;str[i];i++)\n\t{\n\t\tif(str[i]=='(')\n\t\t{ \n\t\t\tfor(end+=1;str[end]!=')'&&str[end]!='\\0';end++);\n\t\t\tif(str[end]=='\\0')                         //remove unbalanced '(' at any position\n\t\t\t{\n\t\t\t\tfor(j=i;str[j]=str[j+1];j++);    \n\t\t\t\ti--;\n\t\t\t}\n\t\t\tif(end<i)\n\t\t\t{\n\t\t\t\tfor(j=end;str[j]=str[j+1];j++);    //remove unbalanced ')' at middle position \n\t\t\t\ti-=1;\n\t\t\t}\n\t\t}\n\t\tif(str[i]==')')                                     //remove unbalanced ')' at last positions\n\t\t        if(i>end || i==0)\n\t\t\t{\n\t\t\t\tfor(j=i;str[j]=str[j+1];j++);\n\t\t\t\ti--;}\n\n\t}\n\tprintf(\"New string is : %s\\n\",str);\n\treturn 0;\n}\n\n"
  },
  {
    "path": "Set 02-02.Remove unbalanced parentheses in a given expression.py",
    "content": "#!/usr/bin/python\n\n#Remove unbalanced parentheses in a given expression\ndef remove(s,index):\n    s=s[:index]+s[index+1:]\n    return s\n    \ndef main():\n         s=raw_input(\"string : \")\n         start,stop=0,0\n         while s[start:].find('(') != -1:\n             \n             index=s[start:].find('(')+start\n            \n             \n             while s[stop:index].find(')')!=-1:\n                 s=remove(s,s[stop:].find(')')+stop)\n                 index-=1\n              \n             if s[stop:].find(')') == -1:\n                 s=remove(s,index)\n              \n                 \n             else:\n                 stop=s[stop:].find(')')+stop+1\n                 start=index+1\n                \n         while s[stop:].find(')')!= -1:\n             s=remove(s,s[stop:].find(')')+stop)\n             \n         print \"formatted string : \",s\nif __name__==\"__main__\":\n        main()\n"
  },
  {
    "path": "Set 02-03.Form a number system with only 3 and 4.c",
    "content": "/* C) Form a number system with only 3 and 4. Find the nth number of the number system.\nEg.) The numbers are: 3, 4, 33, 34, 43, 44, 333, 334, 343, 344, 433, 434, 443, 444, 3333, 3334, 3343, 3344, 3433, 3434, 3443, 3444 ….\n*/\n#include<stdio.h>\nint main(void)\n{\n\tint n,i,m,pos,tmp=1,num_dgs;\n\tprintf(\"Enter the value n  : \");\n\tscanf(\"%d\",&n);\n\tm=n;\n\n\tfor(i=-1;m;m/=2,i++);  //find number of digits\n\tnum_dgs=i;\n\n\tfor(;i;i--)           //find nearest value which is power of two\n\t\ttmp*=2;\n\n\tpos=n-tmp;           \n\tpos++;               //find position of nth  number in the class of num_digits \n                             //eg: 11 is 4th number in 3 digits numbers\n\n\tif(!num_dgs)         //if  n=1 when pos = 1 next step becomes  number of digits  1 and position 0\n\t\tpos=1;\n\n\tif(pos==tmp){       //In cases for example n=7 we have to make number of digits 3 instead of 2 and pos 0\n\t\tpos=0;\n\t\tnum_dgs++;\n\t}\n\n\tprintf(\"The %dth number in sequence is : \",n);\n\tfor(i=num_dgs-1;i>=0;i--)                      //loop in which where 1 is present in bits of pos prints 4 and where 0 print 3\n\t\tif((pos>>i)&1)\n\t\t\tprintf(\"4\");\n\t\telse\n\t\t\tprintf(\"3\");\n\n\tprintf(\"\\n\");\n\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 02-qns",
    "content": "Set 2\n\n1) First Round : Written\n\n40 C output questions. 2 Hours.\n30 1Mark and 10 2Mark questions. IT WAS NOT MCQ. The questions were challenging and covered all C concepts.\n\n2) Second Round : Coding\n\nAround 150 students shortlisted for this round. It was a local machine coding round. A staff will be assigned to a group of 5 students. He made note of the time took for solving each question. There was totally 7 questions and I solved 4 questions and did not complete the 5th question.\n\nA) Alternate sorting: Given an array of integers, rearrange the array in such a way that the first element is first maximum and second element is first minimum.\n\n    Eg.) Input  : {1, 2, 3, 4, 5, 6, 7}\n         Output : {7, 1, 6, 2, 5, 3, 4} \n\nB) Remove unbalanced parentheses in a given expression.\n\n    Eg.) Input  : ((abc)((de))\n         Output : ((abc)(de))  \n\n         Input  : (((ab)\n         Output : (ab) \n\nC) Form a number system with only 3 and 4. Find the nth number of the number system.\nEg.) The numbers are: 3, 4, 33, 34, 43, 44, 333, 334, 343, 344, 433, 434, 443, 444, 3333, 3334, 3343, 3344, 3433, 3434, 3443, 3444 ….\n\nD) Check whether a given mathematical expression is valid.\nZ\n    Eg.) Input  : (a+b)(a*b)\n         Output : Valid\n\n         Input  : (ab)(ab+)\n         Output : Invalid\n\n         Input  : ((a+b)\n         Output : Invalid \n\nI don’t remember the 5th question.\n\n3) Third Round : Advanced Coding\nA matrix game was given with 5 rules. We were asked to implement each of the rules separately.\n\n\n  R3 | -   -   - |\n  R2 | -   -   - |\n  R1 | -   -   - |\n       C1  C2  C3  \n\nEach of the 9 cells can either be empty or filled with an atom. R3, R2, R1 are the rays that originate from the left. C1, C2, C3 are the rays that originate from the bottom of the box.\n\nInput : Position of the atoms and the rays that gets originated from the outside of the box.\n\n  Eg.) 3\n       3 1\n       2 2\n       1 3\n       3\n       R3 C1 C3\n\nOutput  : Print the box. \n\nRule 1:\nA ray that has an atom in its path should print ‘H’ (Hit) If it does not have any atoms in its path, the ray should pass to the other side.\n\n       C1      C3\n  R3 | -   -   - | R3\n  H  | -   X   - |\n  R1 | -   -   - | R1\n       C1  H   C3 \n\nRule 2 & 3:\nA ray that has an atom in its diagonal adjacent position should refract.\n\n  H  | -   -   - |\n  H  | X   -   - |\n  R  | -   X   - |\n       R   H   R  \n\nInput rays: R1, R2, C3\n\n  H  | -   X   - |\n  R2 | -   -   - | C3\n     | -   -   - |\n       R2      C3  \n\nRule 4:\nA ray that has atoms in both of the diagonal adjacent positions should reflect back.\n\n\nInput ray: C2\n   | -   -   - |\n   | X   -   X |\n   | -   -   - |\n         R   \n\nInput ray: R2\n     | -   X   - |\n  R  | -   -   - |\n     | -   X   - | \n\nRule 5:\nThe deflection of rays should happen in the order of the input rays.\n\nInput Rays: R3, R2, C1, C3\n   H | -   X   - |\n  R2 | -   -   - | C3\n     | -   -   - |\n       R2      C3 \n\nThe final task was to implement these rules for dynamic matrix size.\n\n\nInput : no of rows, no of columns\n  Eg.) 4 4 (row & column)\n       2 (No of atoms)\n       4 4 (Position of atom)\n       2 2 (Position of atom)\n       2 (No of rays)\n       R4 C2 (Ray number)\n\n  H  | -   -   -   X |\n     | -   -   -   - |\n     | -   X   -   - |\n     | -   -   -   - |\n           H  \n\nThe final task was very confusing and it had to handle all the cases. There are chances for a ray to end at the starting position if the number of rows and columns are more than 5.\n\n4) Fourth Round : Technical Interview\nBasic questions from hashing, searching, sorting, JVM, OS, Threads. In-depth questions from the projects that I mentioned in my resume. So don’t just add projects that you are not thorough enough to answer all questions.\n\nAnd a simple puzzle : (x-a)(x-b)(x-c)….(x-z) = ? 😀\n\n5) Fifth Round : HR\nGeneral HR questions like why zoho, how do you see yourself after 5 years, why did you choose CS/IT stream, tell me about your leadership skills etc.\n"
  },
  {
    "path": "Set 03-01.program to give the following output.c",
    "content": "/* 1. Write a program to give the following output for the given input\n\nEg 1: Input: a1b10\n       Output: abbbbbbbbbb\nEg: 2: Input: b3c6d15\n          Output: bbbccccccddddddddddddddd\nThe number varies from 1 to 99.*/\n#include<stdlib.h>\n#include<stdio.h>\nint main(void)\n{\nchar *str,last;\nint i,n=0;\nstr=malloc(100);\nprintf(\"Enter the string  : \");\nscanf(\"%s\",str);\n\nfor(i=0;str[i];i++)\n{ \n  if(str[i]<48||str[i]>57)\n  {\n     printf(\"%c\",str[i]);                           //print characters other than integer value and continue the loop\n     continue;\n  }\n  for(last=i;(str[i]>=48&&str[i]<=57)&&str[i];i++)  //Loop to find integer value store it in n\n     n=n*10+(str[i]-48);\n  i--;\n  if(last)                                          //if first value is integer that is last = 0 do nothing simply put n=0 in else part\n  for(n-=1;n;n--)                                   //Loop to print n times the character\n    printf(\"%c\",str[last-1]);            \n  else n=0;\n}\nprintf(\"\\n\");\nfree(str);\nreturn 0;\n}\n \n"
  },
  {
    "path": "Set 03-02.sort the elements in odd positions in descending order and elements in ascending order.c",
    "content": "/* 2. Write a program to sort the elements in odd positions in descending order and elements in ascending order\n\nEg 1: Input : 13,2 4,15,12,10,5\n      Output: 13,2,12,10,5,15,4\nEg 2: Input : 1,2,3,4,5,6,7,8,9\n      Output: 9,2,7,4,5,6,3,8,1 \n*/\n#include<stdio.h>\n#include<stdlib.h>\nint main(void)\n{\n\tint *a,n,i,j;\n\tprintf(\"Array size : \");\n\tscanf(\"%d\",&n);\n\ta=malloc(sizeof(int)*n);\n\tprintf(\"Enter Elements : \");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\"%d\",a+i);\n\tfor(i=0;i<n-1;i++)\n\t\tfor(j=i+1;j<n;j++)\n\t\t{\n\t\t\tif(i%2)\n\t\t\t{\n\t\t\t\tif(a[i]>a[j])\n\t\t\t\t\ta[i]=a[i]+a[j]-(a[j]=a[i]);\n\t\t\t}\n\t\t\telse\n\t\t\t{\n\t\t\t\tif(a[i]<a[j])\n\t\t\t\t\ta[i]=a[i]+a[j]-(a[j]=a[i]);\n\t\t\t}\n\t\t}\n\tprintf(\"Array elements  : \");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"%d \",a[i]);\n        printf(\"\\n\");\n\tfree(a);\n\treturn 0;\n}    \n\n"
  },
  {
    "path": "Set 03-03.program to print the following output for the given input.c",
    "content": "/*3. Write a program to print the following output for the given input. You can assume the string is of odd length\n\nEg 1: Input: 12345\n       Output:\n1       5\n  2   4\n    3\n  2  4\n1      5\nEg 2: Input: geeksforgeeks\n         Output:\ng                         s\n  e                     k\n    e                 e\n      k             e\n        s         g\n          f      r\n             o\n          f     r\n        s         g\n      k             e\n    e                 e\n  e                      k\ng                          s \n*/\n#include<stdio.h>\nint main(void)\n{\n\tchar str[50];\n\tint i,j,n;\n\tprintf(\"Enter the string : \");\n\tscanf(\"%s\",str);\n\tfor(n=0;str[n];n++);\n\tn=n/2+1;\n\tfor(i=1;i<=n;i++)\n\t{\n\t\tfor(j=1;j<i;j++)\n\t\t\tprintf(\" \");\n\t\tfor(j=0;j<2*(n-i)+1;j++)\n\t\t\tif(j==0 || j==2*(n-i))\n\t\t\t\tprintf(\"%c\",str[i-1+j]);\n\t\t\telse\n\t\t\t\tprintf(\" \");\n\t\tprintf(\"\\n\");\n\t}\n\n\tfor(i=2;i<=n;i++)\n\t{\n\t\tfor(j=1;j<=n-i;j++)\n\t\t\tprintf(\" \");\n\t\tfor(j=1;j<=2*i-1;j++)\n\t\t\tif(j==1 || j==2*i-1)\n\t\t\t\tprintf(\"%c\",str[n-i-1+j]);\n\t\t\telse\n\t\t\t\tprintf(\" \");\n\t\tprintf(\"\\n\");\n\t}\n\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 03-05-01.two sorted arrays, merge them such that the elements are not repeated.c",
    "content": "/* 5. Given two sorted arrays, merge them such that the elements are not repeated\n\nEg 1: Input:\n        Array 1: 2,4,5,6,7,9,10,13\n        Array 2: 2,3,4,5,6,7,8,9,11,15\n       Output:\n       Merged array: 2,3,4,5,6,7,8,9,10,11,13,15 */\n\n/* Procedure 1 : Remove repeated terms and saving into third array(lengthy)*/\n#include<stdio.h>\nint main(void)\n{\n\tint a[100],b[100],c[100],i,j,k,m,n;\n\tprintf(\"\\n---------------------------------------------------------------\\n\");\n\tprintf(\"Array  size 1: \");\n\tscanf(\"%d\",&m);\n\tprintf(\"Elements to array 1 : \\n\");\n\tfor(i=0;i<m;i++)\n\t\tscanf(\"%d\",&a[i]);\n\tprintf(\"Array  size 2: \");\n\tscanf(\"%d\",&n);\n\tprintf(\"Elements to array 2 : \\n\");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\"%d\",&b[i]);\n\tprintf(\"\\n---------------------------------------------------------------\\n\");\n\tprintf(\"\\nArray 1 elements :\");\n\tfor(i=0;i<m;i++)\n\t\tprintf(\"%d \",a[i]);\n\tprintf(\"\\nArray 2 elements :\");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"%d \",b[i]);\n\tfor(i=0,j=0,k=0;i<m && j<n;k++)\n\t{\n\t\tif(b[j]<a[i])               //element of array b is smaller then true \n\t\t{\n\t\t\tif(c[k-1]==b[i])    //if previous element in 3rd array is current element then discard it otherwise save it into 3rd array  \n\t\t\t\ti++;\n\t\t\telse{\n\t\t\t\tc[k]=b[j];\n\t\t\t\tj++;\n\t\t\t}\n\t\t}\n\t\telse if(a[i]<b[j])          //element of array a is smaller then true\n\t\t{ \n\t\t\tif(c[k-1]==a[i])    //if previous element in 3rd array is current element then discard it otherwise save it into 3rd array\n\t\t\t\ti++;\n\t\t\telse\n\t\t\t{   \n\t\t\t\tc[k]=a[i];\n\t\t\t\ti++;\n\t\t\t}\n\t\t}\n\t\telse if(a[i]==b[j])        // if both elements in array a and b are  equal then true\n\t\t{  \n\t\t\tif(c[k-1]==a[i]){  //if previous element in 3rd array is current element then discard it otherwise save it into 3rd array\n\t\t\t\ti++;j++;k--;\n\t\t\t}\n\t\t\telse{\n\t\t\t\tc[k]=a[i];\n\t\t\t\ti++;\n\t\t\t\tj++;\n\t\t\t}\n\t\t}\n\n\t}\n\n\twhile(i<m || j<n)            // if m and n are different values the remaing values must store into 3rd array\n\t{\n\t\tif(i<m)              //if i<m true only 1st array elements remaining\n\t\t{\n\t\t\tif(c[k-1]==a[i]) //if previous element in 3rd array is current element then discard it otherwise save it into 3rd array\n\t\t\t\ti++;\n\t\t\telse \n\t\t\t{  \n\t\t\t\tc[k]=a[i];\n\t\t\t\ti++;\n\t\t\t\tk++;\n\t\t\t}\n\t\t}\n\t\tif(j<n)                //if i<m true only 1st array elements remaining\n\t\t{\n\t\t\tif(c[k-1]==b[j]) //if previous element in 3rd array is current element then discard it otherwise save it into 3rd array\n\t\t\t\tj++;\n\t\t\telse\n\t\t\t{\n\t\t\t\tc[k]=b[j];\n\t\t\t\tj++;\n\t\t\t\tk++;\n\t\t\t}\n\t\t}\n\t}\n\tprintf(\"\\nArray 3 elements :\");\n\tfor(i=0;i<k;i++)\n\t\tprintf(\"%d \",c[i]);\n\tprintf(\"\\n\");\n\tprintf(\"\\n---------------------------------------------------------------\\n\");\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 03-05-02.two sorted arrays, merge them such that the elements are not repeated.c",
    "content": "/*5. Given two sorted arrays, merge them such that the elements are not repeated\n\nEg 1: Input:\n        Array 1: 2,4,5,6,7,9,10,13\n        Array 2: 2,3,4,5,6,7,8,9,11,15\n       Output:\n       Merged array: 2,3,4,5,6,7,8,9,10,11,13,15 */\n/*Procedure 2 : Saving both array into 3rd array and removing repeated terms*/\n#include<stdio.h>\nint main(void)\n{\n\tint a[100],b[100],c[100],i,j,k,l,m,n;\n\tprintf(\"\\n---------------------------------------------------------------\\n\");\n\tprintf(\"Array  size 1: \");\n\tscanf(\"%d\",&m);\n\tprintf(\"Elements to array 1 : \\n\");\n\tfor(i=0;i<m;i++)\n\t\tscanf(\"%d\",&a[i]);\n\tprintf(\"Array  size 2: \");\n\tscanf(\"%d\",&n);\n\tprintf(\"Elements to array 2 : \\n\");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\"%d\",&b[i]);\n\tprintf(\"\\n---------------------------------------------------------------\\n\");\n\tprintf(\"\\nArray 1 elements :\");\n\tfor(i=0;i<m;i++)\n\t\tprintf(\"%d \",a[i]);\n\tprintf(\"\\nArray 2 elements :\");\n\tfor(i=0;i<n;i++)\n\t\tprintf(\"%d \",b[i]);\n\tfor(i=0;i<m;c[i]=a[i],i++);    //save array 1 elements to array 3 \n\tfor(j=0;j<n;c[i]=b[j],i++,j++);//save array 2 elements to array 3\n\tl=m+n;                         //l is length of array 3\n\tfor(i=0;i<l;i++)\n\t\tfor(j=i+1;j<l;j++)\n\t\t\tif(c[i]==c[j])   // if repeating terms found in array 3 \n\t\t\t{\n\t\t\t\tfor(k=j;k<l-1;c[k]=c[k+1],k++);   //delete repeating terms \n\t\t\t\tl--;                              //reduce length of array 3 by one\n\t\t\t\tj--;                              //j reduces by one to continue on same jth iteration\n\t\t\t}         \n\n\n\tprintf(\"\\nArray 3 elements :\");\n\tfor(i=0;i<l;i++)\n\t\tprintf(\"%d \",c[i]);\n        printf(\"\\n\");\n        printf(\"\\n---------------------------------------------------------------\\n\");\n        return 0;\n}\n"
  },
  {
    "path": "Set 03-Qns",
    "content": "Set 3\n\nROUND 1: WRITTEN\nAPTITUDE(1 hr and 20 minutes-20 questions): Problems on average, probability, time & distance, alligation&mixture,ratio, HCF & LCM and few a puzzles.\n\nTECHNICAL(45 minutes-10 questions): Output for C questions. Practice questions in geekquiz.com and C output questions in geeksforgeeks.org. Questions in pointers, strings, matrix etc.\n\nNearly 60 students were selected out of 600 candidates. They didn’t select the top 60. They had a cutoff and those who cleared the cutoff were called for the next round\n\nROUND 2: SIMPLE CODING(3 hours)\n\n1. Write a program to give the following output for the given input\n\nEg 1: Input: a1b10\n       Output: abbbbbbbbbb\nEg: 2: Input: b3c6d15\n          Output: bbbccccccddddddddddddddd\nThe number varies from 1 to 99.\n\n2. Write a program to sort the elements in odd positions in descending order and elements in ascending order\n\nEg 1: Input: 13,2 4,15,12,10,5\n        Output: 13,2,12,10,5,15,4\nEg 2: Input: 1,2,3,4,5,6,7,8,9\n        Output: 9,2,7,4,5,6,3,8,1 \n\n3. Write a program to print the following output for the given input. You can assume the string is of odd length\n\nEg 1: Input: 12345\n       Output:\n1       5\n  2   4\n    3\n  2  4\n1      5\nEg 2: Input: geeksforgeeks\n         Output:\ng                         s\n  e                     k\n    e                 e\n      k             e\n        s         g\n          f      r\n             o\n          f     r\n        s         g\n      k             e\n    e                 e\n  e                      k\ng                          s \n\n4. Find if a String2 is substring of String1. If it is, return the index of the first occurrence. else return -1.\n\nEg 1:Input:\n        String 1: test123string\n         String 2: 123\n        Output: 4\nEg 2: Input:\n        String 1: testing12\n        String 2: 1234 \n        Output: -1\n\n5. Given two sorted arrays, merge them such that the elements are not repeated\n\nEg 1: Input:\n        Array 1: 2,4,5,6,7,9,10,13\n        Array 2: 2,3,4,5,6,7,8,9,11,15\n       Output:\n       Merged array: 2,3,4,5,6,7,8,9,10,11,13,15 \n\n6. Using Recursion reverse the string such as\n\nEg 1: Input: one two three\n      Output: three two one\nEg 2: Input: I love india\n      Output: india love I \n\n19 cleared this round and they were called for the next round. The next round took place on the next day\n\nROUND 3: COMPLEX CODING(3 hours)\n\n1) Design a Call taxi booking application\n-There are n number of taxi’s. For simplicity, assume 4. But it should work for any number of taxi’s.\n-The are 6 points(A,B,C,D,E,F)\n-All the points are in a straight line, and each point is 15kms away from the adjacent points.\n-It takes 60 mins to travel from one point to another\n-Each taxi charges Rs.100 minimum for the first 5 kilometers and Rs.10 for the subsequent kilometers.\n-For simplicity, time can be entered as absolute time. Eg: 9hrs, 15hrs etc.\n-All taxi’s are initially stationed at A.\n-When a customer books a Taxi, a free taxi at that point is allocated\n-If no free taxi is available at that point, a free taxi at the nearest point is allocated.\n-If two taxi’s are free at the same point, one with lower earning is allocated\n-Note that the taxi only charges the customer from the pickup point to the drop point. Not the distance it travels from an adjacent point to pickup the customer.\n-If no taxi is free at that time, booking is rejected\n\nDesign modules for\n\n1)    Call taxi booking \nInput 1:\nCustomer ID: 1\nPickup Point: A\nDrop Point: B\nPickup Time: 9\n\nOutput 1:\nTaxi can be allotted.\nTaxi-1 is allotted\n\nInput 2:\nCustomer ID: 2\nPickup Point: B\nDrop Point: D\nPickup Time: 9\n\nOutput 1:\nTaxi can be allotted.\nTaxi-2 is allotted \n\n(Note: Since Taxi-1 would have completed its journey when second booking is done, so Taxi-2 from nearest point A which is free is allocated)\n\n\nInput 3:\nCustomer ID: 3\nPickup Point: B\nDrop Point: C\nPickup Time: 12\n\nOutput 1:\nTaxi can be allotted.\nTaxi-1 is allotted \n\n2) Display the Taxi details\n\n\nTaxi No:    Total Earnings:\nBookingID    CustomerID    From    To    PickupTime    DropTime    Amount\n   \nOutput:\nTaxi-1    Total Earnings: Rs. 400\n\n1    1    A    B     9    10    200\n3    3    B    C    12    13    200\n\nTaxi-2 Total Earnings: Rs. 350\n2    2    B    D    9    11    350 \n\nThese were just sample inputs. It should work for any input that they give.\nThose who finished both the modules within 3 hours and if it worked for all the inputs they give, those candidates were given extra modules to work with.\n\nOnly 9 candidates made it to the next round\n\nROUND 4 : FIRST FACE-TO-FACE(TECHNICAL)\nQuestions were on project, c, oops concepts, DBMS and a few puzzles. They might ask you more on new scenarios relating to your project.\n\nROUND 5: SECOND FACE-TO-FACE(TECHNICAL)\nQuestion were on c, c++, java(like threads, synchronization etc.), Discussion about questions from first, second and third round. He even asked me to solve a few questions from the first round. He gave me a few puzzles to solve\n\nROUND 6: FIRST GENERAL HR\nGeneral questions about my pros and cons and discussion on my resume(be thorough with your resume). She finally asked me if I had any queries.\n\nROUND 7: SECOND GENERAL HR\nShe asked me some family details and gave some scenarios and asked me to what I will do in such situations(like if I am given the power to change 3 things in india, what all will I change) and a few general questions.\n\nI didn’t get direct placement in ZOHO but I got an internship offer. If I perform well in my internship, I will get an offer. Round 3 was the toughest and if you perform exceptionally well and as they expect in that round, you will definitely make it through. Many thanks to geeksforgeeks.org for helping me out in my preparation.\n"
  },
  {
    "path": "Set 08 Qns",
    "content": "\nZoho Interview | Set 8 (Off-Campus Drive for Project Trainee)\n\nFirst Round:\nWritten, pen paper based. The candidate has to write the output for C programs and basic aptitude questions. Total 30 questions. Test duration was of 3 hours.\n\nDescription:\nThere were altogether 20 questions from C programs and 10 questions from logical aptitude. Questions from C programming basically focus on pointers (single pointer and pointer-to-pointer) manipulation, 75% of questions from pointers and rest from arrays and strings manipulation. The questions from logical aptitude was of moderate level difficulty ranging from probability to speed distance time calculations. Both section same marks weightage. I solved 13 from C and 7 from apti.\n\nSecond Round:\nThey shortlisted approx 40 students for the second round. The mail came after 15 days from first round. It was coding round, programming questions on their laptops in groups of 10 students in each room. Test duration was of 3 hours. The test was on 1 Feb 2016.\nDescription:\nThere were 5 questions. The instructor will check the code after solving each question, they may ask for reducing time/ space complexity, be ready for it.\n\n1. Spiral printing.\nO/P\n4444444\n4333334\n4322234\n4321234\n4322234\n4333334\n4444444\n\n2. Sort the array alternately i.e first element should be max value, second min value, third second max, third second min. Eg: arr[] = {1,2,3,4,5,6,7} O/P: {7,1,6,2,5,3,4} Note: no extra space and time complexity should be less;\n\n3. Print all the substring of the given string.\n\n4. Print the numbers which are mismatched from two array. Arr1 = {a b c d e f g h i}\narr2 ={ a b d e e g g i i}, O/P- cd, de, f, g, h, i.\n\n5. Print all possible combinations from the given string.\n\nThe candidate who solved at least four out five are selected for next advance coding round.\n\nThird Round:\nThird round was followed immediately after completion of second round. 50% candidates were selected for this round which consists of advance programming question. One full fledge project consist of 7 modules which has to be done on their laptops. Test duration was of 3 hours. Test was on 1 Feb 2016.\n\nDescription:\nThere was one question for each candidate.\n\n1. Develop a railway project which consist of 7 requirements modules. The modules includes booking of railway tickets, cancellation, route option, fare calculations, allotting/dealloting berth, coach position.\nSome other questions given to students were online billing and shopping. The instructor may ask to change the requirements run time, you should be ready for it.\n\nFourth Round:\nFourth round was followed immediately after completion of third round. It was face to face interview round in which technical and HR questions were asked. Interview duration was of 20 mins for each student. It was on same day.\nDescription:\nThis round consist of face to face interview, they asked questions from Java collection framework and core java. Questions related to course project was also asked. Finally, all the round were finished at 8 pm.\n\n\n\n\n\n"
  },
  {
    "path": "Set 08 Qns2",
    "content": "\nZoho Interview | Set 8 (On-Campus)\n\nROUND 1: Aptitude questions – both Quantitative aptitude and in programming. they were NOT MCQs\n\nROUND 2:\n1. Arrange the numbers in descending order depending on the no. of factors available for each number.\nI/P: {6,8,9}\nO/P: {8,6,9} or {6,8,9}\nReason: factors of 8 (1,2,4,8), factors of 6 (1,2,3,6), factors of 9 (1,3,9).\n\n2. Two strings of equal length are given print the mismatched ones.\nI/P: a b c d e f g h i\n      a b d e e g g i i\nO/P: cd , de //when two char are mismatched they should be printed together.\n      f , g\n      h , i\n\n3. Get a number and check whether its palindrome do not use arrays and string manipulations\nI/P: 5\nO/P: 101-Palindrome\nReason: binary representation of 5 is 101 & it is a palindrome.\nI/P: 10\nO/P: Binary representation of 10 is 1010 –Not a palindrome\n\n4. For any given matrix find the path from the start to the end which gives the maximum sum. Traverse only right or down.\nExample: starting index is 15 (left top) and ending index is 10 (bottom right)\n15 25 30\n45 25 60\n70 75 10\nO/P:15->45->70->75->10 sum is 215\n\n5. [ [‘Lava’ , ‘kusha] ,\n[‘Rama’ , ‘Lava’] ,\n[‘Lava ‘,’Ravanan’] ,\n[‘Abi’ , ‘Lava’] ]\nFirst string is the child & the second string is the parent. Print the no. of grand children available for the\ngiven I/P.\nI/P: Ravanan\nO/P: 2\n"
  },
  {
    "path": "Set 08-01.Spiral printing.c",
    "content": "#include<stdio.h>\nint main(void)\n{\n\tint i,j,n;\n\tprintf(\"Limit   : \");\n\tscanf(\"%d\",&n);\n\tfor(i=n;i>=1;i--)\n\t{\n\t\tfor(j=n;j>=i;j--)\n\t\t\tprintf(\"%d \",j);\n\t\tfor(j=1;j<(2*i)-1;j++)\n\t\t\tprintf(\"%d \",i);\n\t\tfor(j=i+1;j<=n;j++)\n\t\t\tprintf(\"%d \",j);\n\t\tprintf(\"\\n\");\n\t}\n        for(i=2;i<=n;i++)\n\t{\n\t\tfor(j=n;j>=i;j--)\n\t\t\tprintf(\"%d \",j);\n\t\tfor(j=1;j<(2*i)-1;j++)\n\t\t\tprintf(\"%d \",i);\n\t\tfor(j=i+1;j<=n;j++)\n\t\t\tprintf(\"%d \",j);\n\t\tprintf(\"\\n\");\n\t}\n}\n\n"
  },
  {
    "path": "Set 08-02.Two strings of equal length are given print the mismatched ones..c",
    "content": "#include<stdio.h>\nint main(void)\n{\n\tchar a[50],b[50];\n\tint i,j,n,st=0;\n\tprintf(\"Array  size : \");\n\tscanf(\"%d\",&n);\n\tprintf(\"Elements to array :\");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\" %c\",&a[i]);\n\tprintf(\"Elements to array :\");\n\tfor(i=0;i<n;i++)\n\t\tscanf(\" %c\",&b[i]);\n        printf(\"----Output----\\n\");\n\tfor(i=0;i<n;i++)\n\t{\n\t\tif(a[i]!=b[i])\n\t\t{\n\t\t\tfor(j=i;a[j]!=b[j] && j<n;j++)\n\t\t\t\tprintf(\"%c\",a[j]);\n\t\t\tprintf(\" , \");\n\t\t\tfor(;i<j;i++)\n\t\t\t\tprintf(\"%c\",b[i]);\n\t\t\tprintf(\"\\n\");\n\t\t}\n\t}     \n\treturn 0;\n}\n\n"
  },
  {
    "path": "Set 08-03.Get a number and check whether its palindrome do not use arrays and string manipulations.c",
    "content": "#include<stdio.h>\nint main(void)\n{\n\tint n,i,j=0,temp;\n\tprintf(\"Enter a Number : \");\n\tscanf(\"%d\",&n);\n\n\tfor(i=sizeof(int)*8-1;!(n>>i&1);i--);\n\tfor(;j<i;i--,j++)\n\t\tif((n>>i&1)!=(n>>j&1))\n\t\t\tbreak;\n        temp=i;\n\tfor(i+=j;i>=0;printf(\"%d\",n>>i&1),i--);\n\tif(temp<=j)\n\t\tprintf(\" - Palindrome\\n\");\n\telse\n\t\tprintf(\" - Not Palindrome\\n\");\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 12-01.Given two numbers a and b both < 200 we have to find the square numbers which lie between a and b.c",
    "content": "#include<stdio.h>\nint main(void)\n{\nint a,b,i;\nprintf(\"Lower NUmber : \");\nscanf(\"%d\",&a);\nprintf(\"Upper NUmber : \");\nscanf(\"%d\",&b);\nprintf(\"Perfect Square number between %d and %d is : \",a,b);\nfor(i=0;i*i<=a;i++);\nfor(;i*i<b;i++)\nprintf(\"%d \",i*i);\nprintf(\"\\n\");\nreturn 0;\n}\n\n"
  },
  {
    "path": "Set 12-03.Given an array and a threshold value find the output.c",
    "content": "#include<stdio.h>\nint main(void)\n{\nint a[100],i,th,n,count=0,sum,temp;\nprintf(\"Array size : \");\nscanf(\"%d\",&n);\nprintf(\"Elements to Array : \");\nfor(i=0;i<n;i++)\nscanf(\"%d\",&a[i]);\nprintf(\"Array   elements  : \");\nfor(i=0;i<n;i++)\nprintf(\"%d \",a[i]);\nprintf(\"\\nThreshold value   : \");\nscanf(\"%d\",&th);\n\nfor(i=0;i<n;i++)\n{\n  temp=th;\n  sum=0;\n  while(sum!=a[i])\n  {\n    sum+=temp;\n    count++;\n    if(sum>a[i])\n    {\n     sum-=temp;\n     temp--;\n     count--;\n    }\n  }\n}  \nprintf(\"count = %d\\n\",count);\nreturn 0;\n}  \n"
  },
  {
    "path": "Set 12-05.Write a program to print the below pattern.c",
    "content": "#include<stdio.h>\nint main(void)\n{\n\tint n,i,j,k;\n\tprintf(\"Limit : \");\n\tscanf(\"%d\",&n);\n\tfor(i=0;i<n;i++)\n\t{\n\t\tk=i+1;\n\t\tfor(j=0;j<n-i;k+=n-j,j++)\n\t\t\tprintf(\"%2d \",k);\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n"
  },
  {
    "path": "Set 12-06.Given bigger NxN matrix and a smaller MxM matrix print TRUE if the smaller matrix can be found in the bigger matrix else print FALSE.c",
    "content": "\n#include<stdio.h>\nint main(void)\n{\nint a[50][50],b[50][50],r1,c1,r2,c2,i,j,k,l;\nprintf(\"Array 1 size : \");\nscanf(\"%d%d\",&r1,&c1);\nprintf(\"Elements to Array : \");\nfor(i=0;i<r1;i++)\nfor(j=0;j<c1;j++)\nscanf(\"%d\",&a[i][j]);\n\nprintf(\"Array   elements  : \\n\");\nfor(i=0;i<r1;i++)\n{\nfor(j=0;j<c1;j++)\nprintf(\"%d \",a[i][j]);\nprintf(\"\\n\");\n}\nprintf(\"Array 2 size : \");\nscanf(\"%d%d\",&r2,&c2);\n\nprintf(\"Elements to Array : \");\nfor(i=0;i<r2;i++)\nfor(j=0;j<c2;j++)\nscanf(\"%d\",&b[i][j]);\n\nprintf(\"Array   elements  : \\n\");\nfor(i=0;i<r2;i++)\n{\nfor(j=0;j<c2;j++)\nprintf(\"%d \",b[i][j]);\nprintf(\"\\n\");\n}\n\nfor(i=0;i<r1;i++)\nfor(j=0;j<c1;j++)\n{\n if(a[i][j]==b[0][0])\n { \n   for(k=0;k<r2;k++)\n    for(l=0;l<c2;l++)\n      if(a[i+k][j+l]!=b[k][l])\n        goto EXIT;\n    EXIT:\n    if(k==r2 && l==c2)\n    {  \n       printf(\"\\nTRUE : <%d,%d> to <%d,%d>\\n\",i,j,i+k-1,j+l-1);\n       return;\n    }\n }\n}\n printf(\"\\n False\\n\");\n return 0;\n}\n \n\n"
  },
  {
    "path": "Set 12-07.Given two matrices a and b both of size NxN find if matrix a can be transformed to matrix b by rotating it 90deg , 180deg , 270deg if so print TRUE else print FALSE.c",
    "content": "#include<stdio.h>\nint main(void)\n{\nint a[20][20],b[20][20],i,j,r1,c1,r2,c2;\nprintf(\"Row  and  coloumn : \");\nscanf(\"%d %d\",&r1,&c1);\nprintf(\"Elements to array : \");\nfor(i=0;i<r1;i++)\nfor(j=0;j<c1;j++)\nscanf(\"%d\",&a[i][j]);\n\nprintf(\"Row  and  coloumn : \");\nscanf(\"%d %d\",&r2,&c2);\n\nprintf(\"Elements to array : \");\nfor(i=0;i<r2;i++)\nfor(j=0;j<c2;j++)\nscanf(\"%d\",&b[i][j]);\n\n\nprintf(\"Array A   Elements  : \\n\");\nfor(i=0;i<r1;i++)\n{\nfor(j=0;j<c1;j++)\nprintf(\"%d \",a[i][j]);\nprintf(\"\\n\");\n}\n\nprintf(\"Array B  Elements  : \\n\");\nfor(i=0;i<r2;i++)\n{\nfor(j=0;j<c2;j++)\nprintf(\"%d \",b[i][j]);\nprintf(\"\\n\");\n}\n\nif(r2==c1 && c2==r1)\n{\nfor(i=c1-1;i>=0;i--)\nfor(j=0;j<r1;j++)\nif(a[j][i]!=b[c1-1-i][j])\n    goto NEXT1;\nprintf(\"\\nTRUE  (90 Deg)\\n\");\nreturn;\n}\n\nif(r2==r1&&c2==c1)\n{\nfor(i=r1-1;i>=0;i--)\nfor(j=c1-1;j>=0;j--)\nif(a[i][j]!=b[r1-1-i][c1-1-j])\n  goto NEXT2;\nprintf(\"\\nTRUE  (180 Deg)\\n\");\nreturn;\n}\n\nNEXT1:\nif(r2==c1 &&c2==r1)\n{\nfor(i=0;i<c1;i++)\nfor(j=r1-1;j>=0;j--)\nif(a[j][i]!=b[i][r1-1-j])\n   goto NEXT2;\nprintf(\"\\nTRUE (270 Deg)\\n\");\nreturn;\n}\nNEXT2:\nprintf(\"\\nFALSE\\n\");\nreturn 0;\n}\n"
  },
  {
    "path": "Set 12-08.In addition to the above question you have to check if matrix a can be transformed by mirroring vertically or horizontally to matrix b..c",
    "content": "\n#include<stdio.h>\nint main(void)\n{\nint a[20][20],b[20][20],i,j,r1,c1,r2,c2;\nprintf(\"Row  and  coloumn : \");\nscanf(\"%d %d\",&r1,&c1);\nprintf(\"Elements to array : \");\nfor(i=0;i<r1;i++)\nfor(j=0;j<c1;j++)\nscanf(\"%d\",&a[i][j]);\n\nprintf(\"Row  and  coloumn : \");\nscanf(\"%d %d\",&r2,&c2);\n\nprintf(\"Elements to array : \");\nfor(i=0;i<r2;i++)\nfor(j=0;j<c2;j++)\nscanf(\"%d\",&b[i][j]);\n\n\nprintf(\"Array A   Elements  : \\n\");\nfor(i=0;i<r1;i++)\n{\nfor(j=0;j<c1;j++)\nprintf(\"%d \",a[i][j]);\nprintf(\"\\n\");\n}\n\nprintf(\"Array B  Elements  : \\n\");\nfor(i=0;i<r2;i++)\n{\nfor(j=0;j<c2;j++)\nprintf(\"%d \",b[i][j]);\nprintf(\"\\n\");\n}\n\nif(r2==c1 && c2==r1)\n{\nfor(i=c1-1;i>=0;i--)\nfor(j=0;j<r1;j++)\nif(a[j][i]!=b[c1-1-i][j])\n    goto NEXT1;\nprintf(\"\\nTRUE  (90 Deg)\\n\");\nreturn;\n}\n\nif(r2==r1&&c2==c1)\n{\nfor(i=r1-1;i>=0;i--)\nfor(j=c1-1;j>=0;j--)\nif(a[i][j]!=b[r1-1-i][c1-1-j])\n  goto NEXT2;\nprintf(\"\\nTRUE  (180 Deg)\\n\");\nreturn;\n}\n\nNEXT1:\nif(r2==c1 &&c2==r1)\n{\nfor(i=0;i<c1;i++)\nfor(j=r1-1;j>=0;j--)\nif(a[j][i]!=b[i][r1-1-j])\n   goto NEXT4;\nprintf(\"\\nTRUE (270 Deg)\\n\");\nreturn;\n}\nNEXT2:\n\nif(r2==r1&&c2==c1)\n{\nfor(i=0;i<r1;i++)\nfor(j=c1-1;j>=0;j--)\nif(a[i][j]!=b[i][c1-1-j])\n  goto NEXT3;\nprintf(\"\\nTRUE  (Horizontal Mirror)\\n\");\nreturn;\n}\nNEXT3:\n\nif(r2==r1&&c2==c1)\n{\nfor(i=r1-1;i>=0;i--)\nfor(j=0;j<c1;j++)\nif(a[i][j]!=b[r1-1-i][j])\n  goto NEXT4;\nprintf(\"\\nTRUE  (Vertical Mirror)\\n\");\nreturn;\n}\n\nNEXT4:\nprintf(\"\\nFALSE\\n\");\nreturn 0;\n}\n"
  },
  {
    "path": "Set 12-Qns",
    "content": "Zoho Interview Experience | Set 12 (On-Campus)\nRecently ZOHO visited our campus for recruitment and i would like to share my experience.\nThanks to geeksforgeeks which contributed to most of my preparations..\n\nRound 1 : TIME : 2hrs\nI was expecting an aptitude written round .But they gave us 30 flowchart and asked us to go through each flowchart and the questions were based on the flowchart..Questions included finding the o/p of the flowchart , find the missing statement which when added produces the given o/p..If you concentrate and go through the flowchart it should be easy to crack the first round..\n\nRound 2: TIME : 3hrs\nThey shortlisted 150 candidates from the first round..This was a programming round and consisted of straightforward questions..\n\n1.Given two numbers a and b both < 200 we have to find the square numbers which lie between a and b(inclusive)\n\neg) i/p a = 20;b = 100;\n      o/p 25,36,49,64,81,100 \n2.Alternately sort an unsorted array..\n\neg) i/p {5,2,8,7,4,3,9}\n      o/p {9,2,8,3,7,4,5}\n3. Given an array and a threshold value find the o/p\n\neg) i/p {5,8,10,13,6,2};threshold = 3;\n      o/p  count = 17\n      explanation:\nNumber\tparts\t            counts\n5\t            {3,2}                 2\n8                      {3,3,2}              3\n10                    {3,3,3,1}           4\n13                    {3,3,3,3,1}        5\n6                      {3,3}                 2\n2                      {2}                    1 \n4.a. Given two binary numbers add the two numbers in binary form without converting them to decimal value.\n\neg) a = 1010 b = 11001\n      o/p  100011\t            \n  b.The two numbers were given in base n \n\teg) a = 123  b = 13  n = 4\n\t      o/p  202\n5.Write a program to print the below pattern\n\nfor n = 6\n1\t7\t12\t16\t19\t21\n2\t8\t13\t17\t20\n3\t9\t14\t18\t\n4\t10\t15\n5\t11\t\n6 \n6.Given bigger NxN matrix and a smaller MxM matrix print TRUE if the smaller matrix can be found in the bigger matrix else print FALSE\n\n7.Given two matrices a and b both of size NxN find if matrix a can be transformed to matrix b by rotating it 90deg , 180deg , 270deg if so print TRUE else print FALSE\n\n8 In addition to the above question you have to check if matrix a can be transformed by mirroring vertically or horizontally to matrix b.\n\nI solved 7 questions .. Those who solved more than 5 were selected for the next round.\n\n\nRound 3 : TIME : 3hrs\nAbout 50 – 60 were shortlisted .. This round was also a programming round..\nQuestions were based of matrix transformation.Each question was an extension of the previous question in some way..There were 5 questions .. I solved only 2 🙁 But got selected for the next round , I think it was based on my performance on all previous rounds..\n\n\nRound 4:\nWe had both HR and technical HR.\nIn technical HR if they found out that a candidate was not sure on some topic the candidate was asked questions on that topic for the rest of the round even after the candidate said he does not know that concept . In my case it was DBMS 🙁 The round also consisted of 3 puzzles ..they were easy to solve , i solved all of them..\n\nIn HR round we had a friendly chat ..They didn’t want candidates who were willing to quit the job after a few years and go for higher studies( obviously )..I was asked about my family and my hobbies..\n\nAfter round 4 we had to wait for almost 5 hrs before the results were announced..There were 2 packages x and x+y.. I was offered the x package mostly because i didn’t do round 3 well but they said that if we prove ourselves during the internship they would change the package to x+y so looking forward to that 😉\n\nOverall it was a good experience and i really appreciate ZOHO team for the amount of effort they invested in the interview process each round was both challenging and enjoyable in its own way\n"
  }
]