## Algorithm

Problem Name: 1005. Maximize Sum Of Array After K Negations

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.

Example 1:

```Input: nums = [4,2,3], k = 1
Output: 5
Explanation: Choose index 1 and nums becomes [4,-2,3].
```

Example 2:

```Input: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
```

Example 3:

```Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
```

Constraints:

• `1 <= nums.length <= 104`
• `-100 <= nums[i] <= 100`
• `1 <= k <= 104`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public int largestSumAfterKNegations(int[] nums, int k) {
PriorityQueue pq = new PriorityQueue<>();
int sum = 0;
for (int num : nums) {
sum += num;
}
while (k-- > 0) {
sum -= pq.peek();
sum += -1 * pq.peek();
}
return sum;
}
}
``````
Copy The Code &

Input

cmd
nums = [4,2,3], k = 1

Output

cmd
5

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const largestSumAfterKNegations = function(A, K) {
if (A.length === 0) return 0;
A.sort((a, b) => a - b);
let res = 0;
let posIdx;
for (let i = 0, num = 0; i < A.length; i++) {
if (num < K) {
if (A[i] < 0) {
A[i] = -A[i];
} else {
if (posIdx == null) {
posIdx = Math.abs(A[i]) - Math.abs(A[i - 1]) > 0 ? i - 1 : i;
}
A[posIdx] = -A[posIdx];
}
num++;
}
}
res = A.reduce((ac, el) => ac + el, 0);
return res;
};
``````
Copy The Code &

Input

cmd
nums = [4,2,3], k = 1

Output

cmd
5

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def largestSumAfterKNegations(self, A: List[int], K: int) -> int:
heapq.heapify(A)
for _ in range(K):
val = heapq.heappop(A)
heapq.heappush(A, -val)
return sum(A)
``````
Copy The Code &

Input

cmd
nums = [3,-1,0,2], k = 3

Output

cmd
6

### #4 Code Example with C# Programming

```Code - C# Programming```

``````
using System.Collections.Generic;
using System.Linq;

namespace LeetCode
{
public class _1005_MaximizeSumOfArrayAfterKNegations
{
public int LargestSumAfterKNegations(int[] A, int K)
{
var list = new SortedList(A.Length);
var sum = 0;
foreach (var num in A)
{
if (!list.ContainsKey(num))
else
list[num]++;
sum += num;
}

for (int i = 0; i < K; i++)
{
var num = list.First().Key;
if (num == 0) break;

sum -= num * 2;
list[num]--;
if (list[num] == 0)
list.Remove(num);

if (!list.ContainsKey(-num))
else
list[-num]++;
}

return sum;
}
}
}
``````
Copy The Code &

Input

cmd
nums = [3,-1,0,2], k = 3

Output

cmd
6