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3128. Right Triangles 👍

  • Time: $O(mn)$
  • Space: $O(m + n)$
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class Solution {
 public:
  long long numberOfRightTriangles(vector<vector<int>>& grid) {
    long count = 0;
    const int m = grid.size();
    const int n = grid[0].size();
    vector<int> rows(m);
    vector<int> cols(n);

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1) {
          ++rows[i];
          ++cols[j];
        }

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1)
          count += (rows[i] - 1) * (cols[j] - 1);

    return count;
  }
};

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class Solution {
  public long numberOfRightTriangles(int[][] grid) {
    long count = 0;
    final int m = grid.length;
    final int n = grid[0].length;
    int[] rows = new int[m];
    int[] cols = new int[n];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1) {
          ++rows[i];
          ++cols[j];
        }

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1)
          count += (rows[i] - 1) * (cols[j] - 1);

    return count;
  }
}

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class Solution:
  def numberOfRightTriangles(self, grid: list[list[int]]) -> int:
    rows = [0] * len(grid)
    cols = [0] * len(grid[0])

    for i, row in enumerate(grid):
      for j, num in enumerate(row):
        if num == 1:
          rows[i] += 1
          cols[j] += 1

    return sum((rows[i] - 1) * (cols[j] - 1)
               for i, row in enumerate(grid)
               for j, num in enumerate(row)
               if num == 1)