Problem Statement

Given a m x n binary matrix mat (containing only 0s and 1s), return the total number of square submatrices that have all ones.

Example 1:
Input:

mat = [
  [0,1,1,1],
  [1,1,1,1],
  [0,1,1,1]
]

Output: 15
Explanation:

• 10 squares of size 1×1
• 4 squares of size 2×2
• 1 square of size 3×3
  Total = 15

Example 2:
Input:

mat = [
  [1,0,1],
  [1,1,0],
  [1,1,0]
]

Output: 7

Constraints:

• 1 <= m, n <= 300
• mat[i][j] is 0 or 1

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Approach in Plain English

1. We want all squares with all ones in the matrix.
2. Use dynamic programming (DP):
   • Let dp[i][j] = size of the largest square ending at (i, j).
   • If mat[i][j] == 0, dp[i][j] = 0.
   • If mat[i][j] == 1 and not in first row/column, then: dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) Explanation: A square can be extended only if top, left, and top-left cells also form squares.
3. Sum all dp[i][j] values → total squares.

Key idea: DP stores size of largest square ending at each cell, which implicitly counts smaller squares too.

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Beginner-Friendly Dry Run

Take matrix:

mat = [
  [0,1,1],
  [1,1,1],
  [0,1,1]
]

1. Initialize dp matrix same size as mat:

dp = [
  [0,0,0],
  [0,0,0],
  [0,0,0]
]

2. Fill dp:

• Cell (0,0) → mat[0][0] = 0 → dp[0][0] = 0
• Cell (0,1) → mat[0][1] = 1 → dp[0][1] = 1
• Cell (0,2) → mat[0][2] = 1 → dp[0][2] = 1

dp = [
  [0,1,1],
  [0,0,0],
  [0,0,0]
]

• Cell (1,0) → mat[1][0] = 1 → dp[1][0] = 1
• Cell (1,1) → mat[1][1] = 1 → min(dp[0][1], dp[1][0], dp[0][0]) = min(1,1,0) = 0 → dp[1][1] = 1
• Cell (1,2) → mat[1][2] = 1 → min(dp[0][2], dp[1][1], dp[0][1]) = min(1,1,1) = 1 → dp[1][2] = 2

dp = [
  [0,1,1],
  [1,1,2],
  [0,0,0]
]

• Cell (2,0) → mat[2][0] = 0 → dp[2][0] = 0
• Cell (2,1) → mat[2][1] = 1 → min(dp[1][1], dp[2][0], dp[1][0]) = min(1,0,1) = 0 → dp[2][1] = 1
• Cell (2,2) → mat[2][2] = 1 → min(dp[1][2], dp[2][1], dp[1][1]) = min(2,1,1) = 1 → dp[2][2] = 2

dp = [
  [0,1,1],
  [1,1,2],
  [0,1,2]
]

3. Sum all dp values → total squares = 0+1+1 + 1+1+2 + 0+1+2 = 9

---

Textual Approach

• Use DP matrix same size as input.
• For each cell (i,j):
  • If mat[i][j] == 0 → dp[i][j] = 0
  • Else if i==0 or j==0 → dp[i][j] = 1
  • Else → dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
• Sum all dp values → answer.

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Java Code

public class CountSquareSubmatrices {

    public static int countSquares(int[][] mat) {
        int m = mat.length;
        int n = mat[0].length;
        int[][] dp = new int[m][n];
        int total = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (mat[i][j] == 1) {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = 1 + Math.min(dp[i-1][j-1], 
                                         Math.min(dp[i-1][j], dp[i][j-1]));
                    }
                    total += dp[i][j];
                }
            }
        }

        return total;
    }

    public static void main(String[] args) {
        int[][] mat1 = {
            {0,1,1},
            {1,1,1},
            {0,1,1}
        };
        System.out.println(countSquares(mat1)); // Output: 9

        int[][] mat2 = {
            {1,0,1},
            {1,1,0},
            {1,1,0}
        };
        System.out.println(countSquares(mat2)); // Output: 7
    }
}

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Key Points

• DP stores largest square ending at each cell, which implicitly counts smaller squares.
• Time Complexity: O(m*n) → traverse entire matrix once.
• Space Complexity: O(m*n) → DP matrix (can optimize to O(n) if needed).
• Beginner tip: The min function ensures the square can only grow if all three neighbors support it.