What is the pattern for solving the problem "Divide an Array Into Subarrays With Minimum Cost I"?

The key pattern is sorting the array first and then enumerating all possible splits of the array into three subarrays, choosing the split that minimizes the total cost based on the first element of each subarray.

How do I handle multiple possible splits in the array?

You need to evaluate each possible split by calculating the total cost, then choose the split that minimizes the sum of the first elements of the subarrays.

What is the time complexity of this problem?

The time complexity is O(n log n) due to the sorting step, followed by a linear pass to evaluate all possible splits.

Can this problem be solved with a greedy approach?

A greedy approach can be used to some extent, but the optimal solution requires considering all possible splits, not just local decisions, making sorting a crucial step.

What if the array is already sorted?

If the array is already sorted, the problem becomes simpler as the best splits are likely to be closer to each other, but you still need to evaluate possible divisions.

#3010

Easy

auto_awesome数组·排序

LeetCode 题解工作台

将数组分成最小总代价的子数组 I

给你一个长度为 n 的整数数组 nums 。一个数组的代价是它的第一个元素。比方说， [1,2,3] 的代价是 1 ， [3,4,1] 的代价是 3 。你需要将 nums 分成 3 个连续且没有交集的子数组。请你返回这些子数组的最小代价总和。示例 1：输入： num…

数组排序枚举

题目描述

给你一个长度为 n 的整数数组 nums 。

一个数组的代价是它的 第一个 元素。比方说，[1,2,3] 的代价是 1 ，[3,4,1] 的代价是 3 。

你需要将 nums 分成 3 个 连续且没有交集 的子数组。

请你返回这些子数组的最小代价总和。

示例 1：

输入：nums = [1,2,3,12]
输出：6
解释：最佳分割成 3 个子数组的方案是：[1] ，[2] 和 [3,12] ，总代价为 1 + 2 + 3 = 6 。
其他得到 3 个子数组的方案是：
- [1] ，[2,3] 和 [12] ，总代价是 1 + 2 + 12 = 15 。
- [1,2] ，[3] 和 [12] ，总代价是 1 + 3 + 12 = 16 。

示例 2：

输入：nums = [5,4,3]
输出：12
解释：最佳分割成 3 个子数组的方案是：[5] ，[4] 和 [3] ，总代价为 5 + 4 + 3 = 12 。
12 是所有分割方案里的最小总代价。

示例 3：

输入：nums = [10,3,1,1]
输出：12
解释：最佳分割成 3 个子数组的方案是：[10,3] ，[1] 和 [1] ，总代价为 10 + 1 + 1 = 12 。
12 是所有分割方案里的最小总代价。

提示：

3 <= n <= 50
1 <= nums[i] <= 50

lightbulb

解题思路

方法一：遍历找最小值和次小值

我们记数组 $nums$ 的第一个元素为 $a$ ，其余元素中最小的元素为 $b$ ，次小的元素为 $c$ ，那么答案就是 $a+b+c$ 。

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

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class Solution:
    def minimumCost(self, nums: List[int]) -> int:
        a, b, c = nums[0], inf, inf
        for x in nums[1:]:
            if x < b:
                c, b = b, x
            elif x < c:
                c = x
        return a + b + c

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 demonstrates an understanding of sorting-based optimizations for minimizing costs.
question_mark
The candidate shows proficiency in breaking down the problem into smaller, manageable subproblems.
question_mark
The candidate efficiently uses enumeration to handle different possible splits of the array.

warning

常见陷阱

外企场景

error
Forgetting to sort the array before attempting to split it into subarrays, which leads to suboptimal cost results.
error
Not considering all possible splits of the array, leading to missed opportunities for minimizing the total cost.
error
Overcomplicating the problem by attempting to use advanced data structures instead of sorting and simple array manipulations.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Change the number of subarrays to be split into, e.g., 4 or 5 subarrays, adjusting the approach accordingly.
arrow_right_alt
Consider cases where the cost is defined by a different metric, such as the sum of the elements in the subarray instead of the first element.
arrow_right_alt
Limit the possible values of elements in the array to a specific range to explore how the problem scales with constraints.

help

常见问题

外企场景

继续练习

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