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

Code

Thinking Process:

1. 當 iterate 到 matrix[i][j] 時，如果他本身是 1，就可以組成一個 1 x 1 的正方格，而要判斷它是否可以組成一個 size > 1 都是 1 的正方格，就要看他左邊右邊左斜上方的三個格子
2. dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
3. 當 dp[i][j] = n，代表這格可以組成 1 x 1、2 x 2、一路到 n x n 都是 1 的正方形，因此 res += n

Time Complexity: $O(mn)$, Space Complexity: $O(mn)$

class Solution {
public:
    int countSquares(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dp(m, vector<int>(n, 0));
        int res = 0;
        // base case
        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];
            if (j != 0)
                res += dp[0][j];
        }
        // dp: dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
                if (matrix[i][j] == 1) {
                    dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                }
                res += dp[i][j];
            }
        }
        return res;
    }
};

Source

• Count Square Submatrices with All Ones - LeetCode