#3478

Medium

LeetCode Problem Workspace

Choose K Elements With Maximum Sum

Choose K Elements With Maximum Sum involves sorting and selecting elements based on a specific rule, applying array plus sorting techniques.

array sorting heap priority queue

Problem Statement

You are given two integer arrays, nums1 and nums2, both of length n, along with a positive integer k. For each index i from 0 to n - 1, you need to find the result for the corresponding index. To do this, sort nums1 and its corresponding nums2 values together, and perform summations to select the k largest sum values.

Return an array answer where answer[i] represents the result for the index i after selecting the k maximum sums. The goal is to choose elements from nums2 based on the sorted values of nums1, using sorting and heap-based methods for efficient calculation.

Examples

Example 1

Input: nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2

Output: [80,30,0,80,50]

Example 2

Input: nums1 = [2,2,2,2], nums2 = [3,1,2,3], k = 1

Output: [0,0,0,0]

Since all elements in nums1 are equal, no indices satisfy the condition nums1[j] < nums1[i] for any i , resulting in 0 for all positions.

Constraints

n == nums1.length == nums2.length
1 <= n <= 105
1 <= nums1[i], nums2[i] <= 106
1 <= k <= n

Solution Approach

Sort and Match

Start by sorting nums1 and nums2 together based on nums1 values. This allows for efficient pairing of corresponding elements, enabling the identification of the k largest sums by matching elements from nums2 to sorted nums1.

Efficient Summation

After sorting, calculate the sums for each index by considering the first k largest values in the sorted nums1. A priority queue (heap) can be used to track the largest sums efficiently and avoid redundant recalculations.

Return the Result

For each index, based on the pre-calculated sums, return the result corresponding to that index. Ensure that the output is an array of size n, where each value represents the result of summing the k largest elements for that specific index.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sorting step, which is O(n log n). The use of a heap for summation ensures that the operation remains efficient, with space complexity also scaling with O(n) to store intermediate results.

What Interviewers Usually Probe

Checks the candidate's ability to leverage sorting and heap-based techniques.
Assesses understanding of efficient summation strategies and matching.
Evaluates problem-solving under the constraints of time complexity.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the summation process without using heap-based optimization.
Failing to match nums1 and nums2 correctly after sorting.
Incorrectly handling cases where the k elements are not properly selected from nums2.

Follow-up variants

Varying the value of k to test the performance and efficiency of the solution.
Testing with larger arrays and ensuring the algorithm handles large inputs efficiently.
Changing the relationship between nums1 and nums2 to introduce edge cases in the pairing logic.

FAQ

How do I solve the Choose K Elements With Maximum Sum problem?

Start by sorting nums1 and nums2 together based on nums1 values. Then calculate the sum of the k largest sums efficiently using sorting and heap-based methods.

What is the time complexity of the Choose K Elements With Maximum Sum solution?

The time complexity is O(n log n) due to sorting. A heap is used to track the largest sums efficiently, keeping space and time complexity manageable.

Can you explain the relationship between nums1 and nums2 in this problem?

The elements of nums1 and nums2 are paired together after sorting nums1, allowing for efficient calculation of the maximum sums by matching corresponding elements.

Why should I use a heap for this problem?

A heap allows for efficient retrieval and update of the largest sums, reducing the need for repeated sorting or recalculating sums during the process.

What are some potential pitfalls in solving this problem?

Key pitfalls include incorrectly matching nums1 and nums2, failing to use a heap for efficient summation, and miscalculating the k largest sums.

terminal

Solution

Solution 1: Sorting + Priority Queue (Min-Heap)

We can convert the array $\textit{nums1}$ into an array $\textit{arr}$, where each element is a tuple $(x, i)$, representing the value $x$ at index $i$ in $\textit{nums1}$. Then, we sort the array $\textit{arr}$ in ascending order by $x$.

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class Solution:
    def findMaxSum(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:
        arr = [(x, i) for i, x in enumerate(nums1)]
        arr.sort()
        pq = []
        s = j = 0
        n = len(arr)
        ans = [0] * n
        for h, (x, i) in enumerate(arr):
            while j < h and arr[j][0] < x:
                y = nums2[arr[j][1]]
                heappush(pq, y)
                s += y
                if len(pq) > k:
                    s -= heappop(pq)
                j += 1
            ans[i] = s
        return ans

Continue Practicing

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