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, 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

    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)