{"id":26867,"date":"2022-12-20T12:03:41","date_gmt":"2022-12-20T06:33:41","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-how-to-find-contigious-elements-in-a-2d-array-using-c\/"},"modified":"2022-12-20T12:03:41","modified_gmt":"2022-12-20T06:33:41","slug":"solved-how-to-find-contigious-elements-in-a-2d-array-using-c","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-how-to-find-contigious-elements-in-a-2d-array-using-c\/","title":{"rendered":"[Solved] How to find contigious elements in a 2D array using C"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

\n
\n
<\/div>\n
\n
\n

You have a 2D array. The usual way to traverse it, is be double for loop.
\nThe usual order will be this:
\n1st row – 1st column
\n1st row – 2nd column
\n..
\n1st row – last column
\n2nd row – 1st column
\n…
\nlast row – last column<\/p>\n

Assume you are using the order described above, you may want to do something like this:<\/p>\n

Every time you find a land bit, explore the array at the right, upper, right-upper, right-bottom and at the bottom of this position (the other directions are visited before this step) and find the max length of contigius zeros.<\/p>\n

Here is very naive example:<\/p>\n

#include <stdio.h>\n\nint main() {\n  const int N = 4;\n  const int M = 5;\n\n  int a[4][5] =\n  {\n    { 1, 0, 0, 0, 0},\n    { 0, 0, 1, 1, 0},\n    { 0, 1, 0, 0, 0},\n    { 0, 0, 0, 0, 0}\n  };\n\n  \/* A naive implementation, which can optimized.\n   * Some initializations are not needed.\n   *\/\n\n  int i, j, k, z, len;\n  int max_len = -1;\n\n  \/\/ traverse the array in the order described before\n  for(i = 0 ; i < N ; ++i)\n  {\n    for(j = 0 ; j < M ; ++j)\n    {\n      \/* Land bit found! Explore the array in the needed directions. *\/\n      if(a[i][j] == 0) {\n\n        \/\/ remember where we start from\n        k = i;\n        z = j;\n        len = 0;\n        \/\/ search right\n        \/\/ while we are inside the limits of the array\n        \/\/ (here we need to check only the columns, since\n        \/\/ we move to the right)\n        while(z < M) {\n          \/\/ and we find land bits\n          if(a[k][z++] == 0)\n            len++; \/\/ increase the current length\n          else     \/\/ we want contiguous land bits\n            break; \/\/ we found a water bit, so break this loop\n        }\n\n        \/\/ is the current length better than the current found?\n        if(len > max_len)\n          max_len = len; \/\/ yes, so update max_len\n\n\n        \/\/ return back to initial cell we started from\n        k = i;\n        z = j;\n        len = 0;\n        \/\/ search down\n        while(k < N) {\n          if(a[k++][z] == 0)\n            len++;\n          else\n            break;\n        }\n        if(len > max_len)\n          max_len = len;\n\n        k = i;\n        z = j;\n        len = 0;\n        \/\/ search right and down\n        while(k < N && z < M) {\n          if(a[k++][z++] == 0)\n            len++;\n          else\n            break;\n        }\n        if(len > max_len)\n          max_len = len;\n\n        k = i;\n        z = j;\n        len = 0;\n        \/\/ search upper\n        while(k >= 0) {\n          if(a[k--][z] == 0)\n            len++;\n          else\n            break;\n        }\n        if(len > max_len)\n          max_len = len;\n\n        k = i;\n        z = j;\n        len = 0;\n        \/\/ search upper and right\n        while(k >= 0 && z < M) {\n          if(a[k++][z++] == 0)\n            len++;\n          else\n            break;\n        }\n        if(len > max_len)\n          max_len = len;\n      }\n    }\n  }\n\n  printf(\"max_length = %d\\n\", max_len);\n\n  return 0;\n}\n<\/code><\/pre>\n
After understanding the naive approach, try to see where your approach misses. Use a paper and try with smaller examples. This site is not just for debugging. \ud83d\ude42<\/p>\n
[EDIT]<\/p>\n
As Floris mentioned, this example assumes that the land is a straight line, in any direction.<\/p>\n
Based on the answer of Floris, in order to get it work for the U shaped land, you need to modify your countC<\/code> like this:<\/p>\n
else\n{\n    input2[x][y]=2;\n    return 1+countC(input2,x-1,y,r1,c1)+\n        countC(input2,x+1,y,r1,c1)+\n        countC(input2,x,y-1,r1,c1)+\n        countC(input2,x,y+1,r1,c1)+\n        countC(input2,x+1,y+1,r1,c1)+\n        countC(input2,x+1,y-1,r1,c1)+\n        countC(input2,x-1,y+1,r1,c1)+\n        countC(input2,x-1,y-1,r1,c1);\n}\n<\/code><\/pre>\n
as you can see, you would need to go into diagonals too.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            13<\/span> <\/p><\/div>\n<\/div>\n
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solved How to find contigious elements in a 2D array using C <\/p>\n","protected":false},"excerpt":{"rendered":"
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The usual order will be this: 1st row – 1st column 1st row – 2nd column .. 1st row – last column 2nd row – 1st column … last row – last column Assume you are using the … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[361,324],"class_list":["post-26867","post","type-post","status-publish","format-standard","hentry","category-solved","tag-arrays","tag-c"],"yoast_head":"\n[Solved] How to find contigious elements in a 2D array using C - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-how-to-find-contigious-elements-in-a-2d-array-using-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] How to find contigious elements in a 2D array using C - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] You have a 2D array. 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