Leetcode POTD (20 August 2025) – 1277

1277. Count Square Submatrices with All Ones

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

class Solution {
public:
    int m, n;
    int recursiveCountSquares(int i, int j, vector<vector<int>>& matrix, vector<vector<int>>& t){
        if(i>=m||j>=n)
            return 0;

        if(matrix[i][j]==0)
            return 0;

        if(t[i][j] != -1){
            return t[i][j];
        }

        int right = recursiveCountSquares(i,j+1,matrix, t);
        int diagonal = recursiveCountSquares(i+1,j+1, matrix, t);
        int below = recursiveCountSquares(i+1,j, matrix, t);

        return t[i][j] = 1 + min({right, diagonal, below});
    }

    int countSquares(vector<vector<int>>& matrix) {
        m = matrix.size();
        n = matrix[0].size();
        int result = 0;
        vector<vector<int>> t(m+1, vector<int>(n+1, -1));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(matrix[i][j]==1){
                    result += recursiveCountSquares(i,j, matrix, t);
                }
            }
        }
        return result;
    }
};

Complexity:

Space → O(m * n)

Time → O(m * n)