Maximize Sum Of Array After K Negations

Maximize Sum Of Array After K Negations

What is this problem about?

The Maximize Sum Of Array After K Negations problem gives you an integer array and an integer k. You must perform exactly k operations. In one operation, you choose any index i and negate its value (arr[i] = -arr[i]). You can choose the same index multiple times. Your goal is to maximize the sum of the array after performing exactly k negations.

Why is this asked in interviews?

This is a standard Greedy algorithm question that tests your handling of edge cases and simple sorting. Interviewers use it to see if a candidate recognizes the order of operations: you should always negate the most negative numbers first. It also tests if you realize that once all negative numbers are gone, you should just repeatedly negate the smallest positive number to minimize damage to the total sum.

Algorithmic pattern used

This problem strictly follows a Greedy / Sorting pattern.

1. Sort the array in ascending order.
2. Iterate through the array. If you see a negative number and k > 0, negate it and decrement k.
3. After iterating, if k is still greater than 0, it means all numbers are now positive (or zero).
4. Since you must use the remaining k operations, you should repeatedly negate the smallest absolute value in the array. If k is even, negating a number an even number of times does nothing. If k is odd, subtract twice the minimum value from the total sum (once to remove its positive contribution, once to add its negative contribution).

Example explanation

Array: [-4, -1, 0, 3], k = 4.

1. Array is sorted.
2. Index 0 (-4): Negative! Negate to 4. k becomes 3.
3. Index 1 (-1): Negative! Negate to 1. k becomes 2.
4. Index 2 (0): Not negative. Stop negating sequence. Array is now [4, 1, 0, 3]. Sum is 8. We have k = 2 left. We must pick the smallest element to negate repeatedly. The smallest is 0. Negating 0 twice leaves it as 0. Final sum is 8.

If the array was [-4, -1, 2, 3] with k = 3:

• -4 becomes 4, k=2.
• -1 becomes 1, k=1. Array: [4, 1, 2, 3]. Smallest is 1. k=1 (odd). Negate 1 to -1. Sum is $4 - 1 + 2 + 3 = 8$.

Common mistakes candidates make

A very common mistake is forgetting to re-evaluate the "smallest number" after converting all negatives. For example, in [-3, 2], after negating -3, the array is [3, 2]. The smallest number is now 2, not the newly converted 3. If you just blindly negate the last processed index without comparing it to the next element, you will deduct from the wrong number.

Interview preparation tip

When tackling the Maximize Sum Of Array After K Negations coding problem, the cleanest way to find the smallest number after the negative-conversion phase is to keep track of the minimum absolute value seen during the entire loop. This prevents you from having to re-sort or scan the array a second time just to find the target for the leftover odd k operations.