3212. Count Submatrices With Equal Frequency of X and Y

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3212. Count Submatrices With Equal Frequency of X and Y

Description

Given a 2D character matrix grid, where grid[i][j] is either 'X', 'Y', or '.', return the number of submatrices that contain:

  • grid[0][0]
  • an equal frequency of 'X' and 'Y'.
  • at least one 'X'.

Example 1:

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Input: grid = [["X","Y","."],["Y",".","."]]

Output: 3

Explanation:

Example 2:

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Input: grid = [["X","X"],["X","Y"]]

Output: 0

Explanation:

No submatrix has an equal frequency of 'X' and 'Y'.

Example 3:

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Input: grid = [[".","."],[".","."]]

Output: 0

Explanation:

No submatrix has at least one 'X'.

Constraints:

  • 1 <= grid.length, grid[i].length <= 1000
  • grid[i][j] is either 'X', 'Y', or '.'.

Hints/Notes

  • preSum
  • Weekly Contest 405

Solution

Language: C++

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class Solution {
public:
    int numberOfSubmatrices(vector<vector<char>>& grid) {
        int m = grid.size(), n = grid[0].size(), res = 0;
        vector<vector<int>> preSum(m + 1, vector<int>(n + 1, 0));
        vector<vector<int>> count(m + 1, vector<int>(n + 1, 0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int val = 0;
                if (grid[i][j] == 'X') {
                    val = 1;
                } else if (grid[i][j] == 'Y') {
                    val = -1;
                }
                preSum[i + 1][j + 1] =
                    preSum[i][j + 1] + preSum[i + 1][j] - preSum[i][j] + val;
                count[i + 1][j + 1] =
                    count[i][j + 1] + count[i + 1][j] - count[i][j] + abs(val);
                if (preSum[i + 1][j + 1] == 0 && count[i + 1][j + 1] > 0) {
                    res++;
                }
            }
        }
        return res;
    }
};