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

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最新推荐文章于 2023-12-08 13:16:36 发布

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本文介绍了一种结合动态规划与二分查找的算法，用于在一个给定的矩阵中找到最大边长的正方形，其元素之和不超过特定阈值。通过实例演示了如何高效地解决这一问题。

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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：DP+二分

class Solution(object):
    def maxSideLength(self, mat, threshold):
        """
        :type mat: List[List[int]]
        :type threshold: int
        :rtype: int
        """
        import math
        m,n = len(mat),len(mat[0])
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
        for i in range(1,m+1):
            for j in range(1,n+1):
                dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mat[i-1][j-1]

        def ok(res):
            for i in range(res,m+1):
                for j in range(res,n+1):
                    t = dp[i][j]-dp[i][j-res]-dp[i-res][j]+dp[i-res][j-res]
                    if t<=threshold: return True
            return False

        lo,hi = 0,min(m,n)
        res = 0
        while lo<=hi:
            mid=(lo+hi)//2
            if ok(mid):
                res = max(res,mid)
                lo=mid+1
            else:
                hi=mid-1
        return res



s=Solution()
print(s.maxSideLength(mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4))
print(s.maxSideLength(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))
print(s.maxSideLength(mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6))
print(s.maxSideLength(mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184))