Maximize Sum of Array after K Negations Problem & Solution

Given an integer array nums and an integer k, modify the array in the following way: choose an index i and replace nums[i] with -nums[i].

You should apply this process exactly k times. You may choose the same index i multiple times.

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

See the maximize sum of array after k negations problem on LeetCode.

C++ Solution

#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

class Solution {
public:
  int largestSumAfterKNegations(vector<int>& nums, int k) {
    for (int i = 0; i < k; ++i) {
      int lowest = numeric_limits<int>::max();
      int jj = -1;
      for (int j = 0; j < nums.size(); ++j) {
        if (lowest > nums[j]) {
          lowest = nums[j];
          jj = j;
        }
      }

      nums[jj] *= -1;
    }

    int sum = 0;
    for (int i = 0; i < nums.size(); ++i) {
      sum += nums[i];
    }

    return sum;
  }
};

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