**Largest Sum After K Negations - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of integers nums and an integer k. Consider an operation where you pick an element in nums and negate it. Given that you must make exactly k operations, return the maximum resulting sum that can be obtained. Constraints n ≤ 100,000 where n is the length of nums k < 2 ** 31 Example 1 Input nums = [1, 0, -5, -3] k = 4 Output 9 Explanation We can negate -5 and -3 once each and 0 twice to get [1, 0, 5, 3] and its sum is 9.

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int k) {
int l = nums.size();
if (l < 1) return 0;
sort(nums.begin(), nums.end());
int i, sum = 0;
for (i = 0; i < l; i++) {
if (nums[i] >= 0 || k <= 0) break;
nums[i] *= -1;
k--;
sum += nums[i];
}
if (k % 2) {
if (i == l)
sum -= 2 * nums[i - 1];
else if (i == 0)
sum -= 2 * nums[i];
else
sum -= 2 * min(nums[i - 1], nums[i]);
}
for (; i < l; i++) sum += nums[i];
return sum;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
Arrays.sort(nums);
int min = Integer.MAX_VALUE;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] < 0 && k > 0) {
nums[i] = Math.abs(nums[i]);
k--;
} else if (nums[i] == 0)
k = 0;
sum += nums[i];
min = Math.min(nums[i], min);
}
if (k != 0) {
if (k % 2 != 0) {
sum -= min * 2;
}
}
return sum;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
if not nums:
return 0
heap = []
for n in nums:
heapq.heappush(heap, n)
while k > 0:
minElement = heapq.heappop(heap)
if minElement == 0:
break
minElement = minElement * (-1)
heapq.heappush(heap, minElement)
k -= 1
return sum(heap)
```

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