algoadvance

The problem “1005. Maximize Sum Of Array After K Negations” on LeetCode can be described as follows:

Given an array of integers nums and an integer k, you should perform k negations on the elements of the array such that the sum of the array is maximized. Each negation flips the sign of an element from positive to negative or from negative to positive. Return the maximum possible sum of the array after k negations.

Example:

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

Clarifying Questions

  1. Is the array guaranteed to have at least one element?
    • Yes, the array will have at least one element.
  2. Are there any constraints on the values of nums and the value of k?
    • Yes, the elements of nums can range from -10^4 to 10^4 and k can range from 0 to 10^4.
  3. Can we assume the input will always be valid based on these constraints?
    • Yes.

Strategy

Here is the step-by-step strategy to solve this problem:

  1. Sort the Array: Begin by sorting the array. This will help us easily find the smallest (most negative) values to flip.
  2. Flip the Negatives: Iterate through the sorted array, and for each negative value, flip its sign until either we run out of negative values or we reach k flips.
  3. Handle Remaining Flips: If there are still flips left after processing all negative values:
    • If k is even, flipping any number twice ends up with no change in value. Therefore, the sum remains the same.
    • If k is odd, flip the smallest value (positive or negative) to minimize the sum reduction.
  4. Compute the Sum: Finally, compute the sum of the modified array.

Code

def largestSumAfterKNegations(nums, k):
    # Sort the array to handle the smallest (most negative) values first
    nums.sort()
    
    # Flip the sign of negative numbers, if possible
    for i in range(len(nums)):
        if nums[i] < 0 and k > 0:
            nums[i] = -nums[i]
            k -= 1
    
    # If there are flips remaining (k might be even or odd)
    if k > 0:
        # If k is odd, we need to flip the smallest element one more time
        # Sort again to find the smallest element
        nums.sort()
        if k % 2 == 1:
            nums[0] = -nums[0]
    
    # Compute and return the sum of the array
    return sum(nums)

# Example usage:
nums = [4, 2, 3]
k = 1
print(largestSumAfterKNegations(nums, k))  # Output: 5

Time Complexity

Thus, the overall time complexity is (O(n \log n)).

Try our interview co-pilot at AlgoAdvance.com