# 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.

## 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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