LeetCode-in-Java
3212. Count Submatrices With Equal Frequency of X and Y
Medium
Given a 2D character matrix grid, where grid[i][j] is either 'X', 'Y', or '.', return the number of submatrices that contains:
grid[0][0]- an equal frequency of
'X'and'Y'. - at least one
'X'.
Example 1:
Input: grid = [[“X”,”Y”,”.”],[“Y”,”.”,”.”]]
Output: 3
Explanation:
Example 2:
Input: grid = [[“X”,”X”],[“X”,”Y”]]
Output: 0
Explanation:
No submatrix has an equal frequency of 'X' and 'Y'.
Example 3:
Input: grid = [[”.”,”.”],[”.”,”.”]]
Output: 0
Explanation:
No submatrix has at least one 'X'.
Constraints:
1 <= grid.length, grid[i].length <= 1000grid[i][j]is either'X','Y', or'.'.
Solution
public class Solution {
public int numberOfSubmatrices(char[][] grid) {
int n = grid[0].length;
int ans = 0;
int[][] row = new int[n][2];
for (char[] chars : grid) {
int[] count = new int[2];
for (int j = 0; j < n; j++) {
if (chars[j] != '.') {
count[chars[j] - 'X']++;
}
row[j][0] += count[0];
row[j][1] += count[1];
if (row[j][0] > 0 && row[j][0] == row[j][1]) {
ans++;
}
}
}
return ans;
}
}