How do you solve Choose K Elements With Maximum Sum on LeetCode?

Choose K Elements With Maximum Sum (LeetCode #3478) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps to efficiently find valid indices without repeated checks.

What is the brute force solution for Choose K Elements With Maximum Sum?

The brute force approach for Choose K Elements With Maximum Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each element in nums1 and finding all indices in nums1 that are smaller than the current element. We then sum the largest k values from nums2 corresponding to those indices. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Choose K Elements With Maximum Sum?

Choose K Elements With Maximum Sum uses the following concepts: Array, Sorting, Heap (Priority Queue). The recommended patterns to study are: Heap, Sorting, Greedy.

What are the interview tips for Choose K Elements With Maximum Sum?

Always consider edge cases, such as duplicate values in the input arrays. Explain your thought process clearly while coding; this helps the interviewer understand your approach. Practice implementing both brute force and optimized solutions to solidify your understanding.

What are common mistakes when solving Choose K Elements With Maximum Sum?

Not considering the case where all elements in nums1 are equal, leading to incorrect results. Overlooking the need to manage the size of the heap correctly, which can lead to incorrect sums.

#3478

Choose K Elements With Maximum Sum

Medium

ArraySortingHeap (Priority Queue)HeapSortingGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting and a max heap to efficiently track the largest k values from nums2 as we process each element in sorted order of nums1. This reduces the need for repeated sorting.

⚙️

Algorithm

4 steps

1Step 1: Pair nums1 and nums2 into a list of tuples and sort them based on nums1.
2Step 2: Initialize a max heap to keep track of the largest k values from nums2 as we iterate through the sorted list.
3Step 3: For each element in the sorted list, push the corresponding nums2 value into the max heap and, if the heap exceeds size k, pop the smallest element.
4Step 4: The sum of the heap gives the desired result for that index in the original order.

solution.py23 lines

1# Full working Python code
2
3import heapq
4
5def choose_k_elements(nums1, nums2, k):
6    n = len(nums1)
7    indexed = sorted(enumerate(nums1), key=lambda x: x[1])
8    result = [0] * n
9    max_heap = []
10    sum_heap = 0
11    j = 0
12    for i in range(n):
13        while j < n and indexed[j][1] < nums1[i]:
14            heapq.heappush(max_heap, nums2[indexed[j][0]])
15            sum_heap += nums2[indexed[j][0]]
16            if len(max_heap) > k:
17                sum_heap -= heapq.heappop(max_heap)
18            j += 1
19        result[i] = sum_heap
20    return result
21
22# Example usage
23print(choose_k_elements([4,2,1,5,3], [10,20,30,40,50], 2))

ℹ

Complexity note: The complexity is O(n log n) due to sorting the pairs and maintaining a max heap of size k, which allows for efficient retrieval of the largest k elements.

1Sorting helps to efficiently find valid indices without repeated checks.
2Using a max heap allows us to maintain the largest k values dynamically as we process elements.

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