# 1277. Count Square Submatrices with All Ones

## Description

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

## Solutions

• class Solution {
public int countSquares(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
int[][] f = new int[m][n];
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == 0) {
continue;
}
if (i == 0 || j == 0) {
f[i][j] = 1;
} else {
f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i - 1][j], f[i][j - 1])) + 1;
}
ans += f[i][j];
}
}
return ans;
}
}

• class Solution {
public:
int countSquares(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
int ans = 0;
vector<vector<int>> f(m, vector<int>(n));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (matrix[i][j] == 0) continue;
if (i == 0 || j == 0)
f[i][j] = 1;
else
f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
ans += f[i][j];
}
}
return ans;
}
};

• class Solution:
def countSquares(self, matrix: List[List[int]]) -> int:
m, n = len(matrix), len(matrix[0])
f = [[0] * n for _ in range(m)]
ans = 0
for i, row in enumerate(matrix):
for j, v in enumerate(row):
if v == 0:
continue
if i == 0 or j == 0:
f[i][j] = 1
else:
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1
ans += f[i][j]
return ans


• func countSquares(matrix [][]int) int {
m, n, ans := len(matrix), len(matrix[0]), 0
f := make([][]int, m)
for i := range f {
f[i] = make([]int, n)
}
for i, row := range matrix {
for j, v := range row {
if v == 0 {
continue
}
if i == 0 || j == 0 {
f[i][j] = 1
} else {
f[i][j] = min(f[i-1][j-1], min(f[i-1][j], f[i][j-1])) + 1
}
ans += f[i][j]
}
}
return ans
}