# Solving Leetcode Interviews in Seconds with AI: Maximize Sum Of Array After K Negations


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1005" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **Approach:**
    *   Maximize the sum by flipping the most negative numbers first.
    *   If `k` is greater than the number of negative numbers, and `k` is still odd after flipping all negatives, flip the smallest number (which will now be positive) one last time to minimize the reduction in the sum.
    *   If we run out of negative numbers and `k` is even, then all flips are used, and the optimal result is achieved.

*   **Complexity:**
    *   Runtime: O(n log n) due to sorting.
    *   Storage: O(1) (in-place modification). Although sort() can take O(n) space in some implementations, we are aiming for inplace modification here for optimal storage.

	
	# Code
	```python
	def largestSumAfterKNegations(nums, k):
    """
    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.
    """
    nums.sort()
    n = len(nums)
    for i in range(n):
        if nums[i] < 0 and k > 0:
            nums[i] = -nums[i]
            k -= 1

    if k > 0 and k % 2 == 1:
        nums.sort()
        nums[0] = -nums[0]

    return sum(nums)
	```