Problem
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 (Java)
class Solution {
public int countSquares(int[][] matrix) {
int total = 0;
for (int[] ints : matrix) {
total += ints[0];
}
for (int i = 1; i < matrix[0].length; i++) {
total += matrix[0][i];
}
for (int i = 1; i < matrix.length; i++) {
for (int j = 1; j < matrix[0].length; j++) {
if (matrix[i][j] == 1) {
matrix[i][j] =
Math.min(
matrix[i - 1][j - 1],
Math.min(matrix[i - 1][j], matrix[i][j - 1]))
+ 1;
}
total += matrix[i][j];
}
}
return total;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).