Given a m * n matrix of ones and zeros, return how many square submatrices have all ones.
Example 1:
Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is 1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.
Example 2:
Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.
Constraints:
1 <= arr.length <= 300
1 <= arr[0].length <= 300
0 <= arr[i][j] <= 1
Solution:
class Solution {
public int countSquares(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int[][] dp = new int[m][n];
int res = 0;
for (int i = 0; i < m; i ++) {
dp[i][0] = matrix[i][0];
res += dp[i][0];
}
for (int j = 0; j < n; j ++) {
dp[0][j] = matrix[0][j];
res += dp[0][j];
}
// [0, 0] is calculated twice
res -= matrix[0][0];
for (int i = 1; i < m; i ++) {
for (int j = 1; j < n; j ++) {
if (matrix[i][j] == 1) {
dp[i][j] = 1;
dp[i][j] += Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1]));
}
res += dp[i][j];
}
}
// for (int[] arr : dp) System.out.println(Arrays.toString(arr));
return res;
}
}