Introduction

LeetCode 3727 – Maximum Alternating Sum of Squares is an array problem where the goal is to rearrange the elements to obtain the largest possible alternating score.

At first, the formula looks straightforward:

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

However, the challenge is deciding which squared values should occupy the positive positions and which should occupy the negative positions.

A useful observation makes the problem easier: the values added to the score should be as large as possible, while the values subtracted should be as small as possible.

This solution uses sorting to arrange the squared values, then alternates between taking the largest remaining value for a positive position and the smallest remaining value for a negative position.

There is also a small debugging lesson from the submission process: a System.out.println() statement that appears harmless during development can add enough output overhead to contribute to a Time Limit Exceeded (TLE) result.

Problem Statement

Problem Link-: Maximum Alternating Sum of Squares

Given an integer array nums, the elements can be rearranged in any order.

The alternating score is defined as:

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

Return the maximum possible alternating score after rearranging the array.

Example 1

One valid rearrangement is:

Its score is:

2^2-1^2+3^2=4-1+9=12

Example 2

One rearrangement that achieves the maximum is:

The score is:

9-1+4-1+9-4=16

Notice that negative numbers do not remain negative after squaring. For example, ((-3)^2=9), just like (3^2=9).

Understanding the Greedy Idea

Before looking at the code, consider the sign of each position in the alternating sum.

The even indices contribute positively, and the odd indices contribute negatively.

To maximize the total, the positive positions should receive large squared values, while the negative positions should receive small squared values.

For example, suppose the squared values are:

A good arrangement is:

The alternating score becomes:

9-1+4=12

This is better than:

1-4+9=6

The sorting step makes the largest and smallest remaining values easy to access. Two pointers can then select them in the required order.

Java Solution

The following implementation follows the submitted approach: square the elements, sort them, and build a new array by alternating between the largest and smallest remaining values.

Step-by-Step Explanation

1. Square every element

The score uses squares, so each element is converted into its squared value.

For example:

This also explains why the original signs no longer affect the ordering after squaring.

2. Sort the squared values

After sorting:

Now the smallest value is at index 0, and the largest is at the last index.

3. Use two pointers

1. l points to the smallest remaining value.
2. h points to the largest remaining value.
3. v represents the position currently being filled in the new array.

At even positions, the code selects the largest remaining value:

At odd positions, it selects the smallest remaining value:

This creates the desired pattern of large positive contributions and small negative contributions.

4. Calculate the alternating score

The final traversal adds values at even indices and subtracts values at odd indices.

The result is stored in a long, which is important because the sum of squared values can exceed the range of a 32-bit integer.

Dry Run

Consider:

After squaring and sorting:

Initially:

The two pointers select values as follows:

The new array becomes:

Now calculate the alternating score:

Therefore:

The largest squared value contributes positively, while the smallest is subtracted.

A Real Debugging Lesson: Why Did Removing System.out.println() Help?

While testing a solution, it is common to add a print statement to inspect intermediate values:

In this submission, the statement was inside the loop that processes every element.

The interesting part is that the solution received a Time Limit Exceeded result while the print statement was active, but was accepted after the statement was commented out.

Why can that happen?

Console output has a cost

Printing is not the same as assigning a value to a variable. A call to System.out.println() involves output handling, and online judges may capture or process that output.

When the statement executes repeatedly, the accumulated overhead can become significant.

For an array with many elements, a loop such as:

produces up to (n) lines of output.

That work is unnecessary when the final submission only needs to return the answer.

Why commenting it out can matter

Removing the print statement eliminates the repeated console-output operations. This can make a submission faster, particularly when the execution environment imposes overhead on captured output.

However, a TLE is not automatically caused by printing. An inefficient algorithm, repeated string operations, or other unnecessary work can also cause it.

The practical lesson: use print statements while debugging, but remove them before submitting unless the problem explicitly requires output through standard output.

This is a useful reminder that competitive programming performance depends not only on the algorithm but also on the extra work performed inside its loops.

Complexity Analysis

Let (n) be the length of nums.

Time complexity: (O(nlog n))

1. Squaring the elements takes (O(n)).
2. Sorting takes (O(n\log n)).
3. Rearranging the values takes (O(n)).
4. Calculating the alternating sum takes (O(n)).

Sorting dominates the total time complexity.

Auxiliary space: (O(n))

The implementation creates a new array of length (n). The sorting implementation may also use additional internal space.

An Important Implementation Note

The submitted solution works through a somewhat complicated final-element condition:

This handles the remaining element for certain pointer configurations, but the condition is harder to reason about than necessary.

A cleaner implementation can avoid the extra array and special condition by accumulating the score directly while selecting values. It is also important to square into a long if the input constraints might permit values whose squares exceed the int range.

For this problem's stated bounds, the square of an individual value fits in an int, but using long for the score protects the accumulated result.

The broader lesson is to keep pointer conditions simple and test arrays of length one, odd lengths, and even lengths when implementing two-pointer rearrangements.

Interview Takeaway

For array problems involving an alternating sum, identify which positions contribute positively and which contribute negatively before deciding how to sort the data.

The key questions are:

1. Which values should be maximized?
2. Which values should be minimized?
3. Can sorting make those values easy to select?
4. Can two pointers build the required arrangement without unnecessary operations?

This reasoning connects the problem to a broader set of greedy and sorting questions in which the order of elements determines the final score.

Conclusion

LeetCode 3727 – Maximum Alternating Sum of Squares combines array manipulation, sorting, two pointers, and greedy reasoning.

The central idea is to maximize the values added to the score and minimize the values subtracted from it. Squaring the elements first makes the problem about arranging non-negative values, and sorting allows the largest and smallest remaining values to be selected efficiently.

The debugging experience adds one more practical lesson: remove unnecessary console output from performance-sensitive loops before submitting a solution. A small development convenience can create measurable overhead, even when the underlying algorithm is sound.

Together, the greedy observation and the debugging lesson make this problem useful both for algorithm practice and for building better submission habits.