What is the pattern used in the Largest Element in an Array after Merge Operations problem?

The pattern used is greedy choice with invariant validation, ensuring that the best local choices result in the optimal global outcome.

How do you ensure the largest element in the array?

By merging adjacent elements starting from the end of the array, you ensure that the merged sums are as large as possible, leading to the largest final value.

What are the time and space complexities for this problem?

Time complexity depends on the merge operations but can be O(n), and space complexity depends on the array size, with an optimized solution reaching O(1) space.

How does starting from the end of the array improve the solution?

Starting from the end ensures that larger sums are formed early, and adjacent merges combine the largest values possible, which optimizes the final result.

What is a common mistake when solving this problem?

A common mistake is not properly merging adjacent elements from the end of the array or failing to track the largest sum at each step, leading to suboptimal results.

#2789

Medium

auto_awesome贪心·invariant

LeetCode 题解工作台

合并后数组中的最大元素

给你一个下标从 0 开始、由正整数组成的数组 nums 。你可以在数组上执行下述操作任意次：选中一个同时满足 0 和 nums[i] 的下标 i 。将元素 nums[i + 1] 替换为 nums[i] + nums[i + 1] ，并从数组中删除元素 nums[i] 。返回你可以从最终数…

数组贪心

题目描述

给你一个下标从 0 开始、由正整数组成的数组 nums 。

你可以在数组上执行下述操作任意次：

选中一个同时满足 0 <= i < nums.length - 1 和 nums[i] <= nums[i + 1] 的下标 i 。将元素 nums[i + 1] 替换为 nums[i] + nums[i + 1] ，并从数组中删除元素 nums[i] 。

返回你可以从最终数组中获得的最大元素的值。

示例 1：

输入：nums = [2,3,7,9,3]
输出：21
解释：我们可以在数组上执行下述操作：
- 选中 i = 0 ，得到数组 nums = [5,7,9,3] 。
- 选中 i = 1 ，得到数组 nums = [5,16,3] 。
- 选中 i = 0 ，得到数组 nums = [21,3] 。
最终数组中的最大元素是 21 。可以证明我们无法获得更大的元素。

示例 2：

输入：nums = [5,3,3]
输出：11
解释：我们可以在数组上执行下述操作：
- 选中 i = 1 ，得到数组 nums = [5,6] 。
- 选中 i = 0 ，得到数组 nums = [11] 。
最终数组中只有一个元素，即 11 。

提示：

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

lightbulb

解题思路

方法一：倒序合并

根据题目描述，为了最大化合并后的数组中的最大元素，我们应该先合并右侧的元素，使得右侧的元素尽可能大，从而尽可能多地执行合并操作，最终得到最大的元素。

因此，我们可以从右向左遍历数组，对于每个位置 $i$ ，其中 $i \in [0, n - 2]$ ，如果 $nums[i] \leq nums[i + 1]$ ，我们就将 $nums[i]$ 更新为 $nums[i] + nums[i + 1]$ 。这样做，相当于将 $nums[i]$ 与 $nums[i + 1]$ 合并，并且删掉 $nums[i]$ 。

最终，数组中的最大元素就是合并后的数组中的最大元素。

时间复杂度 $O(n)$ ，其中 $n$ 为数组的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def maxArrayValue(self, nums: List[int]) -> int:
        for i in range(len(nums) - 2, -1, -1):
            if nums[i] <= nums[i + 1]:
                nums[i] += nums[i + 1]
        return max(nums)

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
Look for the candidate's ability to identify and implement greedy choices effectively.
question_mark
Candidates who start from the end of the array and merge adjacent elements will demonstrate a good understanding of the problem.
question_mark
Watch for the candidate's explanation of validating each merge operation to avoid suboptimal choices.

warning

常见陷阱

外企场景

error
Failing to correctly merge adjacent elements starting from the end of the array leads to suboptimal results.
error
Not keeping track of the largest possible value during each merge operation can result in missed opportunities to maximize the final element.
error
Overcomplicating the merge strategy or trying to apply complex algorithms that deviate from the greedy approach can slow down the solution.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Varying the number of merges allowed (limiting merges or setting maximum operations)
arrow_right_alt
Using different array sizes and testing edge cases like minimal and maximal input sizes
arrow_right_alt
Exploring the problem with non-integer elements or arrays with a mix of large and small values

help

常见问题

外企场景

继续练习

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