leetcode-167周赛-1292-元素和小于等于阈值的正方形的最大边长

题目描述;

 

 

 

 自己的提交：超时

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m,n = len(mat),len(mat[0])
        res = 0
        for i in range(m):
            for j in range(n):
                i_,j_ = i+res,j+res
                while i_ < m and j_ < n:
                    pre = sum(sum(mat[x][j:j_])for x in range(i,i_))
                    cur = pre + sum(mat[x][j_] for x in range(i,i_+1)) + sum(mat[i_][y] for y in range(j,j_+1)) - mat[i_][j_]
                    if cur <= threshold:
                        res = max(res,j_-j+1)
                    else:
                        break
                    pre = cur
                    i_ += 1
                    j_ += 1
        return res

方法一：前缀和+二分 O(mn*logmin(m,n))

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m, n = len(mat), len(mat[0])
        P = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                P[i][j] = P[i - 1][j] + P[i][j - 1] - P[i - 1][j - 1] + mat[i - 1][j - 1]
        
        def getRect(x1, y1, x2, y2):
            return P[x2][y2] - P[x1 - 1][y2] - P[x2][y1 - 1] + P[x1 - 1][y1 - 1]
        
        l, r, ans = 1, min(m, n), 0
        while l <= r:
            mid = (l + r) // 2
            find = any(getRect(i, j, i + mid - 1, j + mid - 1) <= threshold for i in range(1, m - mid + 2) for j in range(1, n - mid + 2))
            if find:
                ans = mid
                l = mid + 1
            else:
                r = mid - 1
        return ans

 

方法二：O(mn)

class Solution:
    def maxSideLength(self, mat: List[List[int]], threshold: int) -> int:
        m, n = len(mat), len(mat[0])
        P = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                P[i][j] = P[i - 1][j] + P[i][j - 1] - P[i - 1][j - 1] + mat[i - 1][j - 1]
        
        def getRect(x1, y1, x2, y2):
            return P[x2][y2] - P[x1 - 1][y2] - P[x2][y1 - 1] + P[x1 - 1][y1 - 1]
        
        r, ans = min(m, n), 0
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                for c in range(ans + 1, r + 1):
                    if i + c - 1 <= m and j + c - 1 <= n and getRect(i, j, i + c - 1, j + c - 1) <= threshold:
                        ans += 1
                    else:
                        break
        return ans