Leetcode 1005. 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

Difficulty:Easy

Category:

Analyze

Solution

class Solution {
public:
  int largestSumAfterKNegations(vector<int>& A, int K) {
    int sum = 0;
    sort(A.begin(), A.end());

    for (int& a : A) {
      if (K > 0 && a < 0) {
        a = -a;
        K--;
      }
    }

    if (K % 2 == 1) {
      sort(A.begin(), A.end());
      A[0] = -A[0];
    }

    for (int& a : A) sum += a;
    return sum;
  }
};