LeetCode in Kotlin
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
class Solution {
fun numberOfSubmatrices(grid: Array<CharArray>): Int {
val n = grid[0].size
var ans = 0
val row = Array(n) { IntArray(2) }
for (chars in grid) {
val count = IntArray(2)
for (j in 0 until n) {
if (chars[j] != '.') {
count[chars[j].code - 'X'.code]++
}
row[j][0] += count[0]
row[j][1] += count[1]
if (row[j][0] > 0 && row[j][0] == row[j][1]) {
ans++
}
}
}
return ans
}
}