统计全为 1 的正方形子矩阵

/*
 * Solution Explanation:
 * Java Streams are not particularly suited for this type of dynamic programming problem because the solution requires updates in place and dependence on previous values. However, we can still use some functional programming concepts.
 */

import java.util.stream.IntStream;

public class Solution {
    public int countSquares(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int[][] dp = new int[m][n];
        int[] count = {0}; // Use an array to allow lambda to modify
        
        IntStream.range(0, m).forEach(i -> {
            IntStream.range(0, n).forEach(j -> {
                if (matrix[i][j] == 1) {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
                    }
                    count[0] += dp[i][j];
                }
            });
        });
        return count[0];
    }
}