Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:

Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0

Example 3:

Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3

Example 4:

Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

Constraints:

• 1 <= m, n <= 300
• m == mat.length
• n == mat[i].length
• 0 <= mat[i][j] <= 10000
• 0 <= threshold <= 10^5

思路：build 矩阵的prefixsum ，记住建立的时候是 sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + mat[i - 1][j - 1];

利用prefixsum求正方形的时候是：sum[i][j] - sum[i - len][j] - sum[i][j - len] + sum[i - len][j - len]

建立好prefixsum之后，在0，math.min(n,m)之间做binary search， T: n * m * log(min(n,m));

class Solution {
    public int maxSideLength(int[][] mat, int threshold) {
        int n = mat.length;
        int m = mat[0].length;
        int[][] sum = new int[n + 1][m + 1];
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + mat[i - 1][j - 1];
            }
        }
        
        int start = 0; int end = Math.min(n, m);
        while(start + 1 < end) {
            int mid = start + (end - start) / 2;
            if(canhave(mat, mid, threshold, sum)) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if(canhave(mat, end, threshold, sum)) {
            return end;
        }
        if(canhave(mat, start, threshold, sum)) {
            return start;
        }
        return 0;
    }
    
    private boolean canhave(int[][] mat, int len, int threshold, int[][] sum) {
        for(int i = len; i <= mat.length; i++) {
            for(int j = len; j <= mat[0].length; j++) {
                if(sum[i][j] - sum[i - len][j] - sum[i][j - len] + sum[i - len][j - len] <= threshold) {
                    return true;
                }
            }
        }
        return false;
    }
}

思路2：这个比较巧妙，就是在建立prefixsum 矩阵的时候，一起判断，是否有比len更大的合法正方形，有就len++；O(m * n);

class Solution {
    public int maxSideLength(int[][] mat, int threshold) {
        int n = mat.length;
        int m = mat[0].length;
        int[][] sum = new int[n + 1][m + 1];
        int res = 0;
        int len = 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + mat[i - 1][j - 1];
                
                if(i - len >= 0 && j - len >= 0 && 
                  sum[i][j] - sum[i - len][j] - sum[i][j - len] + sum[i - len][j - len] <= threshold) {
                    res = len++;
                }
            }
        }
        return res;
    }
}