3264. Final Array State After K Multiplication Operations I - Detailed Explanation

Problem Statement

You are given an integer array nums and a positive integer k. In one operation, you choose an index i (0-indexed) and update nums[i] to be nums[i] * (-1). You must perform exactly k operations. The goal is to achieve a final state of the array that (in one common formulation) maximizes the total sum. Return the final state of the array after these operations. (If there are multiple valid answers, any one is acceptable.)

Example 1

• Input: nums = [4, 2, 3], k = 1
• Output: [4, -2, 3]
• Explanation:
  There are no negative numbers initially, so flipping any element reduces the sum. To lose as little as possible, you flip the number with the smallest absolute value (2) because flipping 2 to –2 changes the sum by –4, whereas flipping 4 or 3 reduces the sum by more. Hence, the final state is [4, -2, 3].

Example 2

• Input: nums = [3, -1, 0, 2], k = 3
• Output: [3, 1, 0, 2]
• Explanation:
  First, the only negative number (-1) is flipped to become 1 (using one operation). Now, with k = 2 still remaining, you have to perform extra flips. Because there is a 0 in the array, you may flip it arbitrarily (flipping 0 has no effect). (Note that if no 0 existed, you would choose the number with the smallest absolute value; and since two flips cancel each other, an even number of extra operations causes no change while an odd number would force one extra flip.)

Example 3

• Input: nums = [2, -3, -1, 5, -4], k = 2
• Output: [2, 3, -1, 5, 4]
• Explanation:
  The optimal approach is to flip negatives with the largest benefit. Flipping –3 (to 3) yields an increase of 6 in the overall sum; flipping –4 (to 4) yields an increase of 8. After using two operations to flip –3 and –4, the final array state becomes [2, 3, -1, 5, 4].

Constraints:

• 1 ≤ nums.length ≤ 10⁵
• -10⁴ ≤ nums[i] ≤ 10⁴
• 1 ≤ k ≤ 10⁹

Note: If the problem requires you to preserve the original ordering, you must track original indices when processing; otherwise, the final state (even if returned in sorted order) is acceptable.

Hints

1. Priority of Flips:
   Flipping any negative number is beneficial because it converts a subtraction into an addition (increasing the sum by 2 · |value|). So, process negatives first.

2. Handling Extra Operations:
   Once no negative number remains (or if no negatives exist initially), further operations may hurt the sum. In that case, always flip the element with the smallest absolute value. Notice that flipping any number twice returns it to its original state, so if you have an even number of extra operations, the array remains unchanged. Only an odd number of extra operations results in one final flip.

Approaches

Approach 1: Brute Force Simulation (Not Recommended)

One might try to simulate each of the k operations one by one choosing in every step the best element to flip. However, because k can be as high as 10⁹, simulating each operation one at a time would be far too slow.

• Time Complexity: O(k · n) in the worst case, which is infeasible.
• Space Complexity: O(1) extra space (aside from the input).

Approach 2: Greedy Method with Sorting

The greedy method avoids simulating every individual operation by “compressing” the decisions:

1. Flip All Negatives:
   Sort the array (or at least consider the negatives in order of increasing value) so that the most negative numbers are processed first. For every negative number (from the smallest upward), flip it and decrement k until either there are no negatives left or k becomes 0.

2. Extra Operations:
   After flipping negatives, if k still remains, identify the number with the smallest absolute value.

   • If k is even, additional flips cancel out, and the final state does not change.
   • If k is odd, flip that smallest absolute value element one more time.

3. Restoring Order (if required):
   If the final answer must be in the original order, track each element’s original index when sorting. Otherwise, returning the processed (sorted) array state is acceptable.

• Time Complexity:

  • Sorting takes O(n log n)
  • A single pass through the array takes O(n)
  • Overall: O(n log n)

• Space Complexity: O(n) for tracking indices (if required).

Code Solutions

Below are implementations in Python and Java.

Python Implementation

Java Implementation