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Find the K-Sum of an Array

Difficulty: Hard

Problem Description

You are given an integer array nums and a positive integer k. You can choose any subsequence of the array and sum all of its elements together. We define the K-Sum of the array as the kth largest subsequence sum that can be obtained (not necessarily distinct). Return the K-Sum of the array. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Note that the empty subsequence is considered to have a sum of 0.

Key Insights

  • The K-Sum is dependent on the different sums that can be created from subsequences of the array.
  • The maximum number of unique subsequence sums is limited to 2^n, but we only need the k largest sums.
  • Using a max-heap can efficiently keep track of the largest sums while iterating through possible subsequence sums.
  • The challenge lies in efficiently generating and managing the sums to avoid excessive memory usage.

Space and Time Complexity

Time Complexity: O(n * k * log(k)), where n is the number of elements in nums and k is the required K-Sum position.
Space Complexity: O(k), for storing the largest k sums in a max-heap.

Solution

To solve the problem, we can use a max-heap (priority queue) to store the k largest sums we encounter. We start by initializing the heap with the sum of the empty subsequence, which is 0. We then iterate through each number in the array, updating our potential sums by adding the current number to all the sums already in the heap.

For each new sum, we push it into the heap. If the size of the heap exceeds k, we remove the smallest element to ensure we only keep the largest k sums. Finally, the root of the max-heap will be our K-Sum.

Code Solutions

import heapq

def findKSum(nums, k):
    # Max-heap to store the k largest sums
    max_heap = [0]
    
    for num in nums:
        # Generate new sums by adding the current number to existing sums
        new_sums = []
        for current_sum in max_heap:
            new_sum = current_sum + num
            new_sums.append(new_sum)
        
        # Push new sums into the heap
        for new_sum in new_sums:
            heapq.heappush(max_heap, new_sum)
            if len(max_heap) > k:
                heapq.heappop(max_heap)
    
    # The smallest element in the max-heap is the K-Sum
    return heapq.heappop(max_heap)
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