{"id":10039,"date":"2023-04-30T09:52:12","date_gmt":"2023-04-30T16:52:12","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=10039"},"modified":"2023-04-30T09:55:20","modified_gmt":"2023-04-30T16:55:20","slug":"leetcode-2661-first-completely-painted-row-or-column","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-2661-first-completely-painted-row-or-column\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2661. First Completely Painted Row or Column"},"content":{"rendered":"\n

You are given a 0-indexed<\/strong> integer array arr<\/code>, and an m x n<\/code> integer matrix<\/strong> mat<\/code>. arr<\/code> and mat<\/code> both contain all<\/strong> the integers in the range [1, m * n]<\/code>.<\/p>\n\n\n\n

Go through each index i<\/code> in arr<\/code> starting from index 0<\/code> and paint the cell in mat<\/code> containing the integer arr[i]<\/code>.<\/p>\n\n\n\n

Return the smallest index<\/em> i<\/code> at which either a row or a column will be completely painted in<\/em> mat<\/code>.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
\"image<\/figure>\n\n\n\n
Input:<\/strong> arr = [1,3,4,2], mat = [[1,4],[2,3]]\nOutput:<\/strong> 2\nExplanation:<\/strong> The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"image<\/figure>\n\n\n\n
Input:<\/strong> arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]\nOutput:<\/strong> 3\nExplanation:<\/strong> The second column becomes fully painted at arr[3].\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • m == mat.length<\/code><\/li>
  • n = mat[i].length<\/code><\/li>
  • arr.length == m * n<\/code><\/li>
  • 1 <= m, n <= 105<\/sup><\/code><\/li>
  • 1 <= m * n <= 105<\/sup><\/code><\/li>
  • 1 <= arr[i], mat[r][c] <= m * n<\/code><\/li>
  • All the integers of arr<\/code> are unique<\/strong>.<\/li>
  • All the integers of mat<\/code> are unique<\/strong>.<\/li><\/ul>\n\n\n\n
    Solution: Map + Counter<\/strong><\/h2>\n\n\n\n
    Use a map to store the position of each integer.<\/p>\n\n\n\n
    Use row counters and column counters to track how many elements have been painted.<\/p>\n\n\n\n
    Time complexity: O(m*n + m + n)
    Space complexity: O(m*n + m + n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int firstCompleteIndex(vector& arr, vector>& mat) {\n    const int m = mat.size();\n    const int n = mat[0].size();\n    vector rows(m);\n    vector cols(n);\n    vector> pos(m * n + 1);\n    for (int i = 0; i < m; ++i)\n      for (int j = 0; j < n; ++j)\n        pos[mat[i][j]] = {i, j};\n    \n    for (int i = 0; i < arr.size(); ++i) {\n      auto [r, c] = pos[arr[i]];\n      if (++rows[r] == n) return i;\n      if (++cols[c] == m) return i;\n    }\n    return -1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n]. Go through each index i in arr starting from index 0 and paint the…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[70],"tags":[82,216,177],"class_list":["post-10039","post","type-post","status-publish","format-standard","hentry","category-hashtable","tag-hashtable","tag-matrix","tag-medium","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10039","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=10039"}],"version-history":[{"count":4,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10039\/revisions"}],"predecessor-version":[{"id":10043,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/10039\/revisions\/10043"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=10039"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=10039"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=10039"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}