LeetCode #3727 — MEDIUM

Maximum Alternating Sum of Squares

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums. You may rearrange the elements in any order.

The alternating score of an array arr is defined as:

score = arr[0]² - arr[1]² + arr[2]² - arr[3]² + ...

Return an integer denoting the maximum possible alternating score of nums after rearranging its elements.

Example 1:

Input: nums = [1,2,3]

Output: 12

Explanation:

A possible rearrangement for nums is [2,1,3], which gives the maximum alternating score among all possible rearrangements.

The alternating score is calculated as:

score = 2² - 1² + 3² = 4 - 1 + 9 = 12

Example 2:

Input: nums = [1,-1,2,-2,3,-3]

Output: 16

Explanation:

A possible rearrangement for nums is [-3,-1,-2,1,3,2], which gives the maximum alternating score among all possible rearrangements.

The alternating score is calculated as:

score = (-3)² - (-1)² + (-2)² - (1)² + (3)² - (2)² = 9 - 1 + 4 - 1 + 9 - 4 = 16

Constraints:

1 <= nums.length <= 10⁵
-4 * 10⁴ <= nums[i] <= 4 * 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. You may rearrange the elements in any order. The alternating score of an array arr is defined as: score = arr[0]2 - arr[1]2 + arr[2]2 - arr[3]2 + ... Return an integer denoting the maximum possible alternating score of nums after rearranging its elements.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,2,3]

Example 2

[1,-1,2,-2,3,-3]

Step 02

Core Insight

What unlocks the optimal approach

The score uses squares of values. The original signs of <code>nums</code> don't affect the squared terms.
In the alternating sum, even indices contribute positively and odd indices negatively.

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

Upper-end input sizes

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 #3727: Maximum Alternating Sum of Squares
class Solution {
    public long maxAlternatingSum(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            nums[i] *= nums[i];
        }
        Arrays.sort(nums);
        long ans = 0;
        int m = n / 2;
        for (int i = 0; i < m; ++i) {
            ans -= nums[i];
        }
        for (int i = m; i < n; ++i) {
            ans += nums[i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3727: Maximum Alternating Sum of Squares
func maxAlternatingSum(nums []int) (ans int64) {
	for i, x := range nums {
		nums[i] *= x
	}
	slices.Sort(nums)
	m := len(nums) / 2
	for _, x := range nums[:m] {
		ans -= int64(x)
	}
	for _, x := range nums[m:] {
		ans += int64(x)
	}
	return
}

# Accepted solution for LeetCode #3727: Maximum Alternating Sum of Squares
class Solution:
    def maxAlternatingSum(self, nums: List[int]) -> int:
        nums.sort(key=lambda x: x * x)
        n = len(nums)
        s1 = sum(x * x for x in nums[: n // 2])
        s2 = sum(x * x for x in nums[n // 2 :])
        return s2 - s1

// Accepted solution for LeetCode #3727: Maximum Alternating Sum of Squares
// 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 #3727: Maximum Alternating Sum of Squares
// class Solution {
//     public long maxAlternatingSum(int[] nums) {
//         int n = nums.length;
//         for (int i = 0; i < n; ++i) {
//             nums[i] *= nums[i];
//         }
//         Arrays.sort(nums);
//         long ans = 0;
//         int m = n / 2;
//         for (int i = 0; i < m; ++i) {
//             ans -= nums[i];
//         }
//         for (int i = m; i < n; ++i) {
//             ans += nums[i];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3727: Maximum Alternating Sum of Squares
function maxAlternatingSum(nums: number[]): number {
    const n = nums.length;
    for (let i = 0; i < n; i++) {
        nums[i] = nums[i] ** 2;
    }
    nums.sort((a, b) => a - b);
    const m = Math.floor(n / 2);
    let ans = 0;
    for (let i = 0; i < m; i++) {
        ans -= nums[i];
    }
    for (let i = m; i < n; i++) {
        ans += nums[i];
    }
    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 log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.