What is the main algorithmic approach for Split Array Largest Sum?

The problem is best tackled with a combination of binary search and dynamic programming to minimize the largest sum across k subarrays.

Can the problem be solved using only greedy algorithms?

While a greedy approach helps, it must be combined with binary search to efficiently find the minimum largest sum for k subarrays.

How do I optimize the binary search in this problem?

The binary search should focus on finding the smallest possible largest sum by adjusting the boundaries between the maximum element and the total sum of nums.

What are the space and time complexities of the solution?

The time complexity is O(n * log(sum(nums))) and the space complexity is O(n) due to the dynamic programming table used for state transitions.

How can GhostInterview help with this problem?

GhostInterview provides a clear breakdown of how to approach the problem step-by-step, including insights into binary search, dynamic programming, and efficient partitioning.

#410

Hard

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

分割数组的最大值

给定一个非负整数数组 nums 和一个整数 k ，你需要将这个数组分成 k 个非空的连续子数组，使得这 k 个子数组各自和的最大值最小。返回分割后最小的和的最大值。子数组是数组中连续的部分。示例 1：输入： nums = [7,2,5,10,8], k = 2 输出： 18 解释：一…

数组二分查找动态规划贪心前缀和

题目描述

给定一个非负整数数组 nums 和一个整数 k ，你需要将这个数组分成 k 个非空的连续子数组，使得这 k 个子数组各自和的最大值最小。

返回分割后最小的和的最大值。

子数组 是数组中连续的部分。

示例 1：

输入：nums = [7,2,5,10,8], k = 2
输出：18
解释：
一共有四种方法将 nums 分割为 2 个子数组。 
其中最好的方式是将其分为 [7,2,5] 和 [10,8] 。
因为此时这两个子数组各自的和的最大值为18，在所有情况中最小。

示例 2：

输入：nums = [1,2,3,4,5], k = 2
输出：9

示例 3：

输入：nums = [1,4,4], k = 3
输出：4

提示：

1 <= nums.length <= 1000
0 <= nums[i] <= 10⁶
1 <= k <= min(50, nums.length)

lightbulb

解题思路

方法一：二分查找

我们注意到，当子数组的和的最大值越大，子数组的个数越少，当存在一个满足条件的子数组和的最大值时，那么比这个最大值更大的子数组和的最大值一定也满足条件。也就是说，我们可以对子数组和的最大值进行二分查找，找到满足条件的最小值。

我们定义二分查找的左边界 $left = \max(nums)$ ，右边界 $right = sum(nums)$ ，然后对于二分查找的每一步，我们取中间值 $mid = \lfloor \frac{left + right}{2} \rfloor$ ，然后判断是否存在一个分割方式，使得子数组的和的最大值不超过 $mid$ ，如果存在，则说明 $mid$ 可能是满足条件的最小值，因此我们将右边界调整为 $mid$ ，否则我们将左边界调整为 $mid + 1$ 。

我们如何判断是否存在一个分割方式，使得子数组的和的最大值不超过 $mid$ 呢？我们可以使用贪心的方法，从左到右遍历数组，将数组中的元素依次加入到子数组中，如果当前子数组的和大于 $mid$ ，则我们将当前元素加入到下一个子数组中。如果我们能够将数组分割成不超过 $k$ 个子数组，且每个子数组的和的最大值不超过 $mid$ ，则说明 $mid$ 是满足条件的最小值，否则 $mid$ 不是满足条件的最小值。

时间复杂度 $O(n \times \log m)$ ，空间复杂度 $O(1)$ 。其中 $n$ 和 $m$ 分别是数组的长度和数组所有元素的和。

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class Solution:
    def splitArray(self, nums: List[int], k: int) -> int:
        def check(mx):
            s, cnt = inf, 0
            for x in nums:
                s += x
                if s > mx:
                    s = x
                    cnt += 1
            return cnt <= k

        left, right = max(nums), sum(nums)
        return left + bisect_left(range(left, right + 1), True, key=check)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Understanding how binary search can be applied in optimization problems is crucial.
question_mark
Recognizing when to switch between greedy algorithms and dynamic programming is key for this problem.
question_mark
The candidate should be able to efficiently identify the correct partitioning strategy based on dynamic constraints.

warning

常见陷阱

外企场景

error
Misunderstanding the binary search boundaries. Ensure that the lower boundary starts from the maximum element in the array.
error
Incorrect greedy approach leading to too many subarrays. It's important to balance the subarray sums correctly.
error
Failure to optimize dynamic programming transitions efficiently, leading to unnecessary computations and increased complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Allowing k to be a dynamic value, where it changes based on additional constraints or conditions.
arrow_right_alt
Adapting the problem to handle negative numbers in nums, which may change how partitions are handled.
arrow_right_alt
Optimizing the solution further using a more efficient greedy strategy or advanced dynamic programming techniques.

help

常见问题

外企场景

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