Maximize Sum Of Array After K Negations

Given an array A of integers, we must modify the array in the following way: we choose an i and replace A[i] with -A[i], and we repeat this process K times in total. (We may choose the same index i multiple times.)

Return the largest possible sum of the array after modifying it in this way.

Example 1:

Input: A = [4,2,3], K = 1
Output: 5
Explanation: Choose indices (1,) and A becomes [4,-2,3].

Example 2:

Input: A = [3,-1,0,2], K = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and A becomes [3,1,0,2].

Example 3:

Input: A = [2,-3,-1,5,-4], K = 2
Output: 13
Explanation: Choose indices (1, 4) and A becomes [2,3,-1,5,4].

Note:

1. 1 <= A.length <= 10000

2. 1 <= K <= 10000

3. -100 <= A[i] <= 100

class Solution {
    public int largestSumAfterKNegations(int[] A, int K) {
        Arrays.sort(A);
        // flipping all negative numbers (count<k)
        for (int i = 0; K > 0 && i < A.length && A[i] < 0; ++i, --K)
            A[i] = -A[i];
        int res = 0, min = Integer.MAX_VALUE;
        for (int a : A) {
            res += a;
            min = Math.min(min, a);
        }
        // remaining k is even then we can just flip same number again and again and it has no effect
        // if k is odd then we need to reduce to min number ( as it's sign will be flipped )
        // 2 is in multipliplicatio because min has to be removed with original sign twice from original sum
        return res - (K % 2) * min * 2;
    }
}