What is the key to solving the 'Maximum Element-Sum of a Complete Subset of Indices' problem?

The key is understanding prime factorization and identifying subsets where every pair of indices' product results in a perfect square.

How can I optimize my solution for large inputs in the 'Maximum Element-Sum of a Complete Subset of Indices' problem?

Use dynamic programming or memoization to efficiently find valid subsets, and apply number theory to avoid unnecessary computations.

What is the mathematical principle that makes two indices' product a perfect square?

Two indices' product is a perfect square if the product of their prime factors has even exponents for every prime factor.

What are the common pitfalls when solving the 'Maximum Element-Sum of a Complete Subset of Indices' problem?

Common pitfalls include not utilizing prime factorization efficiently, failing to optimize the solution, and using brute force methods for large inputs.

Can I solve the 'Maximum Element-Sum of a Complete Subset of Indices' problem without using number theory?

Number theory techniques, especially prime factorization, are critical for solving this problem efficiently and correctly, so avoiding them is not recommended.

#2862

Hard

auto_awesome数组·数学

LeetCode 题解工作台

完全子集的最大元素和

给你一个下标从 1 开始、由 n 个整数组成的数组。你需要从 nums 选择一个完全集，其中每对元素下标的乘积都是一个完全平方数，例如选择 a i 和 a j ， i * j 一定是完全平方数。返回完全子集所能取到的最大元素和。示例 1：输入： nums = [8,7,3,5,…

数组数学数论

题目描述

给你一个下标从 1 开始、由 n 个整数组成的数组。你需要从 nums 选择一个 完全集，其中每对元素下标的乘积都是一个完全平方数，例如选择 a_i 和 a_j ，i * j 一定是完全平方数。

返回 完全子集 所能取到的 最大元素和 。

示例 1：

输入：nums = [8,7,3,5,7,2,4,9]

输出：16

解释：

我们选择下标为 2 和 8 的元素，并且 2 * 8 是一个完全平方数。

示例 2：

输入：nums = [8,10,3,8,1,13,7,9,4]

输出：20

解释：

我们选择下标为 1, 4, 9 的元素。1 * 4, 1 * 9, 4 * 9 是完全平方数。

提示：

1 <= n == nums.length <= 10⁴
1 <= nums[i] <= 10⁹

lightbulb

解题思路

方法一：枚举

我们注意到，如果一个数字可以表示成 $k \times j^2$ 的形式，那么所有该形式的数字的 $k$ 是相同的。

因此，我们可以在 $[1,..n]$ 范围内枚举 $k$ ，然后从 $1$ 开始枚举 $j$ ，每一次累加 $nums[k \times j^2 - 1]$ 的值到 $t$ 中，直到 $k \times j^2 > n$ 。此时更新答案为 $ans = \max(ans, t)$ 。

最后返回答案 $ans$ 即可。

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

1

2

3

4

5

6

7

8

9

10

11

12

13

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        n = len(nums)
        ans = 0
        for k in range(1, n + 1):
            t = 0
            j = 1
            while k * j * j <= n:
                t += nums[k * j * j - 1]
                j += 1
            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
Look for understanding of prime factorization and perfect square properties.
question_mark
Check if the candidate can apply number theory efficiently in subset problems.
question_mark
Evaluate the ability to optimize solutions with dynamic programming or other techniques.

warning

常见陷阱

外企场景

error
Overlooking the importance of prime factorization for identifying perfect square pairs.
error
Failing to optimize the solution for large input sizes (n up to 10^4).
error
Choosing a brute force approach without considering number theory for efficiency.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Extend the problem by finding the subset with the minimum sum instead of the maximum sum.
arrow_right_alt
Consider a variant where the perfect square condition is replaced with a different mathematical property (e.g., perfect cube).
arrow_right_alt
Modify the problem to allow subsets that are not necessarily contiguous but still respect the perfect square condition.

help

常见问题

外企场景

继续练习

#2818 操作使得分最大 #2761 和等于目标值的质数对 #2748 美丽下标对的数目