Maximize Sum Of Array After K Negations

The task requires you to modify an integer array in such a way that you can replace an element at an index ‘i’ with its negative counterpart. You have to do this operation ‘k’ times, and you can pick the same index ‘i’ multiple times. The goal is to maximize the sum of the array after performing these operations.

To solve this problem, follow these steps:

1. Sort the given array in ascending order. This ensures that the negative numbers (if any) will be at the beginning of the array.

2. Iterate through the sorted array and flip the sign of negative numbers while decrementing the count of ‘k’. Once you encounter a non-negative number or exhaust ‘k’, stop.

3. If there are still some operations left (k > 0) and ‘k’ is an even number, then no more operations are needed because flipping a number twice will return it to its original state.

4. If ‘k’ is an odd number, then we need to flip the smallest number in the array once more to minimize the impact on the total sum. The smallest number can be either the last number we flipped (if there were any negative numbers), or it could be the next non-negative number.

5. Finally, calculate and return the sum of the modified array.

Python solution:

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class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        nums.sort()

        for i in range(len(nums)):
            if nums[i] < 0 and k > 0:
                nums[i] = -nums[i]
                k -= 1

            if k == 0:
                break

        if k % 2 == 1:
            nums[nums.index(min(nums))] *= -1

        return sum(nums)

This solution works within the given constraints and has a time complexity of O(n log n) due to sorting the array, where n is the length of the array. The space complexity is O(1) as we’re modifying the input array in place and not using any additional data structures.