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 <= A.length <= 10000
1 <= K <= 10000
-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;
}
}