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

---

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        nums=sorted(nums)
        i=0
        for i in range(len(nums)):
            if nums[i]<0:
                nums[i]=-nums[i]
                k-=1
                if k==0:
                    return sum(nums)
        
        if k%2==0:
            return sum(nums)
        else:
            nums=sorted(nums)
            return sum(nums[1::])-nums[0]