1005. Maximize Sum Of Array After K Negations

Problem Link

Description

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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

Solution

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Python3

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        neg = []
        pos = []
        
        for x in nums:
            if x < 0:
                neg.append(x)
            else:
                pos.append(x)
        
        pos.sort()
        
        if k >= len(neg):
            k -= len(neg)
            nums = sorted(abs(x) for x in nums)
            
            if k % 2 == 1:
                return -nums[0] + sum(nums[1:])
            else:
                return sum(nums)
        else:
            neg.sort()
            
            return sum(pos) - sum(neg[:k]) + sum(neg[k:])