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

原创

是念 2022-08-11 17:45:46 博主文章分类：python ©著作权

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Description

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

分析

题目的意思是：求出构成的正方形且小于阈值的最大正方形，这道题目我也不会，看懂了也不会做，后面发现可以先用dp数组把以当前位置结尾的最大正方形的和求出来，然后再一个一个的遍历找最大的正方形。求和的代码为：

dp[i][j]=mat[i-1][j-1]+dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]

存放的是以i，j结尾的最大矩形的和。

res=dp[i][j]-dp[i][j-k]-dp[i-k][j]+dp[i-k][j-k]

res存放的是以k为边构成的矩形的和。

代码

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m=len(mat)
        n=len(mat[0])
        least=min(m,n)
        dp=[[0]*(n+1) for i in range(m+1)]
        for i in range(1,m+1):
            for j in range(1,n+1):
                dp[i][j]=mat[i-1][j-1]+dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]
                
        res=0
        for k in range(least,-1,-1):
            for i in range(k,m+1):
                for j in range(k,n+1):
                    res=dp[i][j]-dp[i][j-k]-dp[i-k][j]+dp[i-k][j-k]
                    if(res<=threshold):
                        return k
        return 0