Solution Explanation: Subsequence of Size K With the Largest Even Sum

This problem asks to find the largest even sum of a subsequence of length k from a given array nums. The solution uses a greedy approach combined with sorting.

Approach:

1. Sort: The array nums is sorted in ascending order. This allows us to easily select the largest k elements as a starting point for our subsequence.

2. Initial Sum: The sum of the largest k elements is calculated. If this sum is already even, it's the answer, and we're done.

3. Odd Sum Handling: If the initial sum is odd, we need to adjust the subsequence to make the sum even. This is done by considering two possible replacements:

   • Replace smallest even with largest odd: We find the smallest even number among the largest k elements and the largest odd number among the remaining n-k elements. Replacing the smallest even with the largest odd will maintain the subsequence length and make the sum even.

   • Replace smallest odd with largest even: Similarly, we find the smallest odd number among the largest k elements and the largest even number among the remaining n-k elements. Replacing the smallest odd with the largest even will achieve the same result.

4. Maximum Even Sum: The algorithm compares the initial sum (if even), the sum after the first replacement, and the sum after the second replacement. The largest of these even sums is returned. If none of these scenarios produce an even sum, -1 is returned.

Time Complexity Analysis:

• Sorting the array takes O(n log n) time, where n is the length of nums.
• All other operations (finding the minimum and maximum values, calculating sums, and comparisons) take O(n) time in the worst case.

Therefore, the overall time complexity is dominated by the sorting step, making it O(n log n).

Space Complexity Analysis:

• The space complexity is primarily determined by the sorting algorithm used. In-place sorting algorithms like merge sort or heapsort have a space complexity of O(log n) due to the recursive call stack.

Therefore, the space complexity is O(log n).

Code Examples (Python):

import sys
 
def largestEvenSum(nums: list[int], k: int) -> int:
    nums.sort()
    ans = sum(nums[-k:])  # Sum of largest k elements
    if ans % 2 == 0:
        return ans
 
    n = len(nums)
    mx1 = mx2 = -sys.maxsize  # Initialize with negative infinity
    for x in nums[:n - k]:
        if x % 2:
            mx1 = max(mx1, x)
        else:
            mx2 = max(mx2, x)
 
    mi1 = mi2 = sys.maxsize  # Initialize with positive infinity
    for x in nums[-k:][::-1]: # Iterate in reverse
        if x % 2:
            mi2 = min(mi2, x)
        else:
            mi1 = min(mi1, x)
 
    ans = max(ans - mi1 + mx1 if mx1 != -sys.maxsize and mi1 != sys.maxsize else -sys.maxsize,
              ans - mi2 + mx2 if mx2 != -sys.maxsize and mi2 != sys.maxsize else -sys.maxsize,
              -1)
 
    return ans
 

This improved Python code handles edge cases more robustly (e.g., when no suitable even or odd numbers are found for replacement) by using sys.maxsize and explicitly checking for valid replacements. The other language examples follow a similar logic and time/space complexity.