1292.Maximum-Side-Length-of-a-Square-with-Sum-Less-than-or-Equal-to-Threshold

1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold

题目地址

https://leetcode.com/problems/maximum-side-length-of-a-square-with-sum-less-than-or-equal-to-threshold/

题目描述

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

代码

Approach #1

Time: O(1) && Space: O(1)

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

    for (int k = Math.min(m, n) - 1; k > 0; k--) {
      for (int i = 1; i+k <= m; i++) {
        for (int j = 1; j+k <= n; j++) {
          if (prefixSum[i+k][j+k] - prefixSum[i-1][j+k] 
              - prefixSum[i+k][j-1] + prefixSum[i-1][j-1] <= threshold) {
            return k+1;
          }
        }
      }
    }
    return 0;
  }
}