{"id":8555,"date":"2021-08-09T20:24:57","date_gmt":"2021-08-10T03:24:57","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8555"},"modified":"2021-08-09T20:26:18","modified_gmt":"2021-08-10T03:26:18","slug":"leetcode-1895-largest-magic-square","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/geometry\/leetcode-1895-largest-magic-square\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1895. Largest Magic Square"},"content":{"rendered":"\n

A k x k<\/code> magic square<\/strong> is a k x k<\/code> grid filled with integers such that every row sum, every column sum, and both diagonal sums are all equal<\/strong>. The integers in the magic square do not have to be distinct<\/strong>. Every 1 x 1<\/code> grid is trivially a magic square<\/strong>.<\/p>\n\n\n\n

Given an m x n<\/code> integer grid<\/code>, return the size<\/strong> (i.e., the side length <\/em>k<\/code>) of the largest magic square<\/strong> that can be found within this grid<\/em>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[7,1,4,5,6],[2,5,1,6,4],[1,5,4,3,2],[1,2,7,3,4]]\nOutput:<\/strong> 3\nExplanation:<\/strong> The largest magic square has a size of 3.\nEvery row sum, column sum, and diagonal sum of this magic square is equal to 12.\n- Row sums: 5+1+6 = 5+4+3 = 2+7+3 = 12\n- Column sums: 5+5+2 = 1+4+7 = 6+3+3 = 12\n- Diagonal sums: 5+4+3 = 6+4+2 = 12\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[5,1,3,1],[9,3,3,1],[1,3,3,8]]\nOutput:<\/strong> 2\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • m == grid.length<\/code><\/li>
  • n == grid[i].length<\/code><\/li>
  • 1 <= m, n <= 50<\/code><\/li>
  • 1 <= grid[i][j] <= 106<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: Brute Force w\/ Prefix Sum<\/strong><\/h2>\n\n\n\n
    Compute the prefix sum for each row and each column.<\/p>\n\n\n\n
    And check all possible squares.<\/p>\n\n\n\n
    Time complexity: O(m*n*min(m,n)2<\/sup>)
    Space complexity: O(m*n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int largestMagicSquare(vector>& grid) {\n    const int m = grid.size();\n    const int n = grid[0].size();\n    vector> rows(m, vector(n + 1));\n    vector> cols(n, vector(m + 1));\n    for (int i = 0; i < m; ++i)\n      for (int j = 0; j < n; ++j) {\n        rows[i][j + 1] = rows[i][j] + grid[i][j];\n        cols[j][i + 1] = cols[j][i] + grid[i][j];\n      }\n    for (int k = min(m, n); k >= 2; --k)\n      for (int i = 0; i + k <= m; ++i)\n        for (int j = 0; j + k <= n; ++j) {\n          vector s;\n          for (int r = 0; r < k; ++r) \n            s.push_back(rows[i + r][j + k] - rows[i + r][j]);\n          for (int c = 0; c < k; ++c)\n            s.push_back(cols[j + c][i + k] - cols[j + c][i]);\n          int d1 = 0;\n          int d2 = 0;\n          for (int d = 0; d < k; ++d) {\n            d1 += grid[i + d][j + d];\n            d2 += grid[i + d][j + k - d - 1];\n          }\n          s.push_back(d1);\n          s.push_back(d2);          \n          if (count(begin(s), end(s), s[0]) == s.size()) return k;\n        }\n    return 1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    A k x k magic square is a k x k grid filled with integers such that every row sum, every column sum, and both diagonal sums are all equal. The…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[177,200],"class_list":["post-8555","post","type-post","status-publish","format-standard","hentry","category-geometry","tag-medium","tag-prefix-sum","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8555","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=8555"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8555\/revisions"}],"predecessor-version":[{"id":8557,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8555\/revisions\/8557"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8555"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8555"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}