What is the optimal approach for solving the 'Minimize Maximum of Array' problem?

The optimal approach involves using binary search to narrow down the possible maximum values combined with greedy operations to validate each candidate maximum.

How can binary search be applied in the 'Minimize Maximum of Array' problem?

Binary search is applied by testing different candidate maximum values and checking if it's feasible to achieve them using a series of operations.

What is the time complexity of the solution?

The time complexity is O(n log M), where n is the length of the array and M is the range between the minimum and maximum values in nums.

Why is a prefix sum important in this problem?

Prefix sums are used to efficiently compute the cumulative effect of operations on the array, allowing fast feasibility checks during binary search iterations.

What are some potential pitfalls when solving this problem?

Common pitfalls include mismanaging binary search bounds, failing to validate candidate maximum values properly, and not using an efficient method like prefix sums for large arrays.

#2439

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

最小化数组中的最大值

给你一个下标从 0 开始的数组 nums ，它含有 n 个非负整数。每一步操作中，你需要：选择一个满足 1 的整数 i ，且 nums[i] > 0 。将 nums[i] 减 1 。将 nums[i - 1] 加 1 。你可以对数组执行任意次上述操作，请你返回可以得到的 nums 数组…

数组二分查找动态规划贪心前缀和

题目描述

给你一个下标从 0 开始的数组 nums ，它含有 n 个非负整数。

每一步操作中，你需要：

选择一个满足 1 <= i < n 的整数 i ，且 nums[i] > 0 。
将 nums[i] 减 1 。
将 nums[i - 1] 加 1 。

你可以对数组执行任意次上述操作，请你返回可以得到的 nums 数组中 最大值 最小为多少。

示例 1：

输入：nums = [3,7,1,6]
输出：5
解释：
一串最优操作是：
1. 选择 i = 1 ，nums 变为 [4,6,1,6] 。
2. 选择 i = 3 ，nums 变为 [4,6,2,5] 。
3. 选择 i = 1 ，nums 变为 [5,5,2,5] 。
nums 中最大值为 5 。无法得到比 5 更小的最大值。
所以我们返回 5 。

示例 2：

输入：nums = [10,1]
输出：10
解释：
最优解是不改动 nums ，10 是最大值，所以返回 10 。

提示：

n == nums.length
2 <= n <= 10⁵
0 <= nums[i] <= 10⁹

lightbulb

解题思路

方法一：二分查找

最小化数组的最大值，容易想到二分查找。我们二分枚举数组的最大值 $mx$ ，找到一个满足题目要求的、且值最小的 $mx$ 即可。

时间复杂度 $O(n \times \log M)$ ，其中 $n$ 为数组的长度，而 $M$ 为数组中的最大值。

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class Solution:
    def minimizeArrayValue(self, nums: List[int]) -> int:
        def check(mx):
            d = 0
            for x in nums[:0:-1]:
                d = max(0, d + x - mx)
            return nums[0] + d <= mx

        left, right = 0, max(nums)
        while left < right:
            mid = (left + right) >> 1
            if check(mid):
                right = mid
            else:
                left = mid + 1
        return left

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
The candidate is familiar with binary search applications in optimization problems.
question_mark
The candidate demonstrates understanding of greedy algorithms in combination with binary search.
question_mark
The candidate efficiently manages the prefix sum to validate the feasibility of each operation.

warning

常见陷阱

外企场景

error
Misunderstanding the operation limits may lead to incorrect binary search bounds.
error
Failing to efficiently validate each candidate maximum value within the binary search process.
error
Not using prefix sums or an alternative efficient method for checking feasibility in large arrays.

swap_horiz

进阶变体

外企场景

arrow_right_alt
In some variants, the number of allowed operations may be restricted, requiring additional constraints handling.
arrow_right_alt
The problem may be extended to multi-dimensional arrays, increasing the complexity of both binary search and validation.
arrow_right_alt
Instead of greedy operations, some variants may involve dynamic programming or different optimization methods for determining the result.

help

常见问题

外企场景

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