What is the most efficient way to solve the 'Maximum Strength of a Group' problem?

The most efficient way is to use dynamic programming with state transitions, optimizing subset evaluations to avoid redundant calculations.

How does backtracking help in solving the 'Maximum Strength of a Group' problem?

Backtracking helps by pruning unnecessary calculations, speeding up the solution and improving overall efficiency, especially for larger inputs.

What is the time complexity of the 'Maximum Strength of a Group' problem?

The time complexity depends on the approach chosen, but can range from O(2^n) for brute force to O(n^2) or better with dynamic programming optimizations.

How can dynamic programming help in the 'Maximum Strength of a Group' problem?

Dynamic programming helps by storing and transitioning between previously computed subset strengths, avoiding the need to re-evaluate subsets multiple times.

What are common mistakes when solving 'Maximum Strength of a Group'?

Common mistakes include failing to prune the search space, not managing state transitions efficiently, and missing optimal subsets due to incorrect subset evaluations.

#2708

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

一个小组的最大实力值

给你一个下标从 0 开始的整数数组 nums ，它表示一个班级中所有学生在一次考试中的成绩。老师想选出一部分同学组成一个非空小组，且这个小组的实力值最大，如果这个小组里的学生下标为 i 0 , i 1 , i 2 , ... , i k ，那么这个小组的实力值定义为 nums[i 0 ] *…

数组动态规划回溯贪心位运算

题目描述

给你一个下标从 0 开始的整数数组 nums ，它表示一个班级中所有学生在一次考试中的成绩。老师想选出一部分同学组成一个非空小组，且这个小组的 实力值 最大，如果这个小组里的学生下标为 i₀, i₁, i₂, ... , i_k ，那么这个小组的实力值定义为 nums[i₀] * nums[i₁] * nums[i₂] * ... * nums[i_k] 。

请你返回老师创建的小组能得到的最大实力值为多少。

示例 1：

输入：nums = [3,-1,-5,2,5,-9]
输出：1350
解释：一种构成最大实力值小组的方案是选择下标为 [0,2,3,4,5] 的学生。实力值为 3 * (-5) * 2 * 5 * (-9) = 1350 ，这是可以得到的最大实力值。

示例 2：

输入：nums = [-4,-5,-4]
输出：20
解释：选择下标为 [0, 1] 的学生。得到的实力值为 20 。我们没法得到更大的实力值。

提示：

1 <= nums.length <= 13
-9 <= nums[i] <= 9

lightbulb

解题思路

方法一：二进制枚举

题目实际上是求所有子集的乘积的最大值，由于数组长度不超过 $13$ ，我们可以考虑使用二进制枚举的方法。

我们在 $[1, 2^n)$ 的范围内枚举所有的子集，对于每个子集，我们计算其乘积，最后返回最大值即可。

时间复杂度 $O(2^n \times n)$ ，其中 $n$ 是数组的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def maxStrength(self, nums: List[int]) -> int:
        ans = -inf
        for i in range(1, 1 << len(nums)):
            t = 1
            for j, x in enumerate(nums):
                if i >> j & 1:
                    t *= x
            ans = max(ans, t)
        return ans

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 understanding of dynamic programming by implementing state transitions effectively.
question_mark
The candidate optimizes subset evaluation to avoid redundant calculations, showing a good grasp of time complexity management.
question_mark
The candidate exhibits proficiency in using backtracking to prune the search space and improve algorithm efficiency.

warning

常见陷阱

外企场景

error
Failing to prune redundant calculations, which leads to inefficient solutions and long execution times.
error
Not properly managing state transitions, causing incorrect results or excessive computations.
error
Forgetting to account for the full range of possible subsets, leading to missing optimal solutions.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider larger input sizes and how the algorithm could scale with more students (larger nums array).
arrow_right_alt
Optimize the space complexity by reducing the amount of state tracking used during dynamic programming transitions.
arrow_right_alt
Adapt the solution for problems with different constraints, such as varying the score range or array length.

help

常见问题

外企场景

继续练习

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