What is the optimal approach for solving Max Sum of Rectangle No Larger Than K?

The optimal approach involves binary search on the sum space combined with prefix sum arrays and ordered sets to efficiently track the maximum valid rectangle sum.

How can I optimize the space complexity for this problem?

Using prefix sums and ordered sets helps in minimizing redundant space usage, allowing for efficient tracking of sums without recalculating them multiple times.

What is the time complexity of solving the Max Sum of Rectangle No Larger Than K problem?

The time complexity depends on the matrix size and the binary search, which typically leads to O(n^2 log k) complexity in the worst case.

What are common mistakes when solving this problem?

Common mistakes include failing to implement binary search correctly, neglecting to optimize with prefix sums, or improperly handling edge cases like negative values in the matrix.

How does binary search improve the solution to this problem?

Binary search helps by narrowing the possible sum space, reducing the number of potential rectangle sums that need to be checked, which optimizes performance.

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LeetCode 题解工作台

矩形区域不超过 K 的最大数值和

给你一个 m x n 的矩阵 matrix 和一个整数 k ，找出并返回矩阵内部矩形区域的不超过 k 的最大数值和。题目数据保证总会存在一个数值和不超过 k 的矩形区域。示例 1：输入： matrix = [[1,0,1],[0,-2,3]], k = 2 输出： 2 解释：蓝色边框圈出来的…

数组二分查找矩阵前缀和有序集合

题目描述

给你一个 m x n 的矩阵 matrix 和一个整数 k ，找出并返回矩阵内部矩形区域的不超过 k 的最大数值和。

题目数据保证总会存在一个数值和不超过 k 的矩形区域。

示例 1：

输入：matrix = [[1,0,1],[0,-2,3]], k = 2
输出：2
解释：蓝色边框圈出来的矩形区域 [[0, 1], [-2, 3]] 的数值和是 2，且 2 是不超过 k 的最大数字（k = 2）。

示例 2：

输入：matrix = [[2,2,-1]], k = 3
输出：3

提示：

m == matrix.length
n == matrix[i].length
1 <= m, n <= 100
-100 <= matrix[i][j] <= 100
-10⁵ <= k <= 10⁵

进阶：如果行数远大于列数，该如何设计解决方案？

lightbulb

解题思路

方法一：枚举边界 + 有序集合

我们可以枚举矩形的上下边界 $i$ 和 $j$ ，然后计算出该边界内每列的元素和，记录在数组 $nums$ 中，问题转化为如何在数组 $nums$ 中寻找不超过 $k$ 的最大子数组和。

我们可以使用有序集合来快速寻找小于等于 $x$ 的最大值，从而得到最大子数组和不超过 $k$ 的子数组。

时间复杂度 $O(m^2 \times n \times \log n)$ ，空间复杂度 $O(n)$ 。

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class Solution:
    def maxSumSubmatrix(self, matrix: List[List[int]], k: int) -> int:
        m, n = len(matrix), len(matrix[0])
        ans = -inf
        for i in range(m):
            nums = [0] * n
            for j in range(i, m):
                for h in range(n):
                    nums[h] += matrix[j][h]
                s = 0
                ts = SortedSet([0])
                for x in nums:
                    s += x
                    p = ts.bisect_left(s - k)
                    if p != len(ts):
                        ans = max(ans, s - ts[p])
                    ts.add(s)
        return ans

speed

复杂度分析

指标	值
时间	complexity depends on the binary search and matrix operations. In the worst case, the solution may involve iterating through rows and columns, leading to a time complexity of O(n^2 log k). Space complexity is influenced by the need to store prefix sums and ordered sets, resulting in O(n^2) space.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Ability to efficiently handle a dynamic programming problem with constraints.
question_mark
Skill in optimizing matrix-based solutions with space and time complexity considerations.
question_mark
Proficiency in using binary search and ordered data structures for algorithmic optimization.

warning

常见陷阱

外企场景

error
Failing to correctly implement the binary search to narrow down valid sums.
error
Neglecting to optimize sum calculations using prefix sums, leading to inefficient solutions.
error
Improper handling of edge cases, such as when k is very small or the matrix contains negative values.

swap_horiz

进阶变体

外企场景

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Max Sum of Submatrix with Exact Sum: Instead of maximizing the sum, the goal is to find a submatrix with a sum equal to k.
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Max Sum of Rectangle No Larger Than K with Multiple Queries: Handling multiple k values for the same matrix and optimizing for each query.
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Max Sum of Subrectangle in a 2D Matrix with Larger k: An extension where k is larger, requiring different search space handling.

help

常见问题

外企场景

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