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1727. Largest Submatrix With Rearrangements 👍

  • Time: $O(mn\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int largestSubmatrix(vector<vector<int>>& matrix) {
    const int n = matrix[0].size();
    int ans = 0;
    vector<int> hist(n);

    for (const auto& row : matrix) {
      // accumulate the histogram if possible
      for (int i = 0; i < n; ++i)
        hist[i] = row[i] == 0 ? 0 : hist[i] + 1;

      // get sorted histogram
      vector<int> sortedHist(hist);
      sort(begin(sortedHist), end(sortedHist));

      // greedily calculate the answer
      for (int i = 0; i < n; ++i)
        ans = max(ans, sortedHist[i] * (n - i));
    }

    return ans;
  }
};
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class Solution {
  public int largestSubmatrix(int[][] matrix) {
    final int n = matrix[0].length;
    int ans = 0;
    int[] hist = new int[n];

    for (int[] row : matrix) {
      // accumulate the histogram if possible
      for (int i = 0; i < n; ++i)
        hist[i] = row[i] == 0 ? 0 : hist[i] + 1;

      // get sorted histogram
      int[] sortedHist = hist.clone();
      Arrays.sort(sortedHist);

      // greedily calculate the answer
      for (int i = 0; i < n; ++i)
        ans = Math.max(ans, sortedHist[i] * (n - i));
    }

    return ans;
  }
}
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class Solution:
  def largestSubmatrix(self, matrix: List[List[int]]) -> int:
    ans = 0
    hist = [0] * len(matrix[0])

    for row in matrix:
      # accumulate the histogram if possible
      for i, num in enumerate(row):
        hist[i] = 0 if num == 0 else hist[i] + 1

      # get sorted histogram
      sortedHist = sorted(hist)

      # greedily calculate the answer
      for i, h in enumerate(sortedHist):
        ans = max(ans, h * (len(row) - i))

    return ans