{"id":7194,"date":"2020-08-02T12:18:39","date_gmt":"2020-08-02T19:18:39","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7194"},"modified":"2020-08-02T12:19:08","modified_gmt":"2020-08-02T19:19:08","slug":"leetcode-1536-minimum-swaps-to-arrange-a-binary-grid","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/uncategorized\/leetcode-1536-minimum-swaps-to-arrange-a-binary-grid\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1536. Minimum Swaps to Arrange a Binary Grid"},"content":{"rendered":"\n

Given an n x n<\/code> binary grid<\/code>, in one step you can choose two adjacent rows<\/strong> of the grid and swap them.<\/p>\n\n\n\n

A grid is said to be valid<\/strong> if all the cells above the main diagonal are zeros<\/strong>.<\/p>\n\n\n\n

Return the minimum number of steps<\/em> needed to make the grid valid, or -1<\/strong> if the grid cannot be valid.<\/p>\n\n\n\n

The main diagonal of a grid is the diagonal that starts at cell (1, 1)<\/code> and ends at cell (n, n)<\/code>.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[0,0,1],[1,1,0],[1,0,0]]\nOutput:<\/strong> 3\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]\nOutput:<\/strong> -1\nExplanation:<\/strong> All rows are similar, swaps have no effect on the grid.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[1,0,0],[1,1,0],[1,1,1]]\nOutput:<\/strong> 0\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == grid.length<\/code><\/li>
  • n == grid[i].length<\/code><\/li>
  • 1 <= n <= 200<\/code><\/li>
  • grid[i][j]<\/code> is 0<\/code> or 1<\/code><\/li><\/ul>\n\n\n\n
    Solution: Bubble Sort<\/strong><\/h2>\n\n\n\n
    Counting how many tailing zeros each row has.
    Then input
    [0, 0, 1]
    [1, 1, 0]
    [1, 0, 0]
    becomes [0, 1, 2]
    For i-th row, it needs n – i – 1 tailing zeros.
    We need to find the first<\/strong> row that has at least<\/strong> n – i – 1 tailing zeros and bubbling it up to the i-th row. This process is very similar to bubble sort.
    [0, 1, 2<\/strong>] -> [0, 2<\/strong>, 1] -> [2<\/strong>, 0, 1]
    [2, 0, 1<\/strong>] -> [2, 1<\/strong>, 0]
    Total 3 swaps.
    <\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  int minSwaps(vector>& grid) {\n    const int n = grid.size();\n    vector zeros;\n    for (const auto& row : grid) {\n      int zero = 0;\n      for (int i = n - 1; i >= 0 && row[i] == 0; --i) \n        ++zero;\n      zeros.push_back(zero);\n    }\n    int ans = 0;\n    for (int i = 0; i < n; ++i) {\n      int j = i;\n      int z = n - i - 1; \/\/ the i-th needs n - i - 1 zeros\n      \/\/ Find first row with at least z tailing zeros.\n      while (j < n && zeros[j] < z) ++j;\n      if (j == n) return -1;\n      while (j > i) {\n        zeros[j] = zeros[j - 1];\n        --j;\n        ++ans;\n      }\n    }\n    return ans;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them. A grid is said to be valid if all the cells…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-7194","post","type-post","status-publish","format-standard","hentry","category-uncategorized","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7194","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=7194"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7194\/revisions"}],"predecessor-version":[{"id":7196,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/7194\/revisions\/7196"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=7194"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=7194"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=7194"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}