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;
}
};