LeetCode #2862 — HARD

Maximum Element-Sum of a Complete Subset of Indices

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 1-indexed array nums. Your task is to select a complete subset from nums where every pair of selected indices multiplied is a perfect square,. i. e. if you select a_i and a_j, i * j must be a perfect square.

Return the sum of the complete subset with the maximum sum.

Example 1:

Input: nums = [8,7,3,5,7,2,4,9]

Output: 16

Explanation:

We select elements at indices 2 and 8 and 2 * 8 is a perfect square.

Example 2:

Input: nums = [8,10,3,8,1,13,7,9,4]

Output: 20

Explanation:

We select elements at indices 1, 4, and 9. 1 * 4, 1 * 9, 4 * 9 are perfect squares.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a 1-indexed array nums. Your task is to select a complete subset from nums where every pair of selected indices multiplied is a perfect square,. i. e. if you select ai and aj, i * j must be a perfect square. Return the sum of the complete subset with the maximum sum.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[8,7,3,5,7,2,4,9]

Example 2

[8,10,3,8,1,13,7,9,4]

Core Insight

What unlocks the optimal approach

Define P(x) as the product of primes p with odd exponents in x's factorization. Examples: For <code>x = 18</code>, factorization <code>21 × 32</code>, P(18) = 2; for <code>x = 45</code>, factorization <code>32 × 51</code>, P(45) = 5; for <code>x = 50</code>, factorization <code>21 × 52</code>, P(50) = 2; for <code>x = 210</code>, factorization <code>21 × 31 × 51 × 71</code>, P(210) = 210.
If <code>P(i) = P(j)</code>, <code>nums[i]</code> and <code>nums[j]</code> can be grouped together.
Pick the group with the largest sum.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
class Solution {
    public long maximumSum(List<Integer> nums) {
        long ans = 0;
        int n = nums.size();
        boolean[] used = new boolean[n + 1];
        int bound = (int) Math.floor(Math.sqrt(n));
        int[] squares = new int[bound + 1];
        for (int i = 1; i <= bound + 1; i++) {
            squares[i - 1] = i * i;
        }
        for (int i = 1; i <= n; i++) {
            long res = 0;
            int idx = 0;
            int curr = i * squares[idx];
            while (curr <= n) {
                res += nums.get(curr - 1);
                curr = i * squares[++idx];
            }
            ans = Math.max(ans, res);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
func maximumSum(nums []int) (ans int64) {
	n := len(nums)
	for k := 1; k <= n; k++ {
		var t int64
		for j := 1; k*j*j <= n; j++ {
			t += int64(nums[k*j*j-1])
		}
		ans = max(ans, t)
	}
	return
}

# Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
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

// Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
// class Solution {
//     public long maximumSum(List<Integer> nums) {
//         long ans = 0;
//         int n = nums.size();
//         boolean[] used = new boolean[n + 1];
//         int bound = (int) Math.floor(Math.sqrt(n));
//         int[] squares = new int[bound + 1];
//         for (int i = 1; i <= bound + 1; i++) {
//             squares[i - 1] = i * i;
//         }
//         for (int i = 1; i <= n; i++) {
//             long res = 0;
//             int idx = 0;
//             int curr = i * squares[idx];
//             while (curr <= n) {
//                 res += nums.get(curr - 1);
//                 curr = i * squares[++idx];
//             }
//             ans = Math.max(ans, res);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2862: Maximum Element-Sum of a Complete Subset of Indices
function maximumSum(nums: number[]): number {
    let ans = 0;
    const n = nums.length;
    for (let k = 1; k <= n; ++k) {
        let t = 0;
        for (let j = 1; k * j * j <= n; ++j) {
            t += nums[k * j * j - 1];
        }
        ans = Math.max(ans, t);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.