Is Maximize Points After Choosing K Tasks easy or hard?

Maximize Points After Choosing K Tasks is generally classified as a medium difficulty problem. The challenge is recognizing the greedy pattern and combining sorting with a heap to manage dynamically available choices. Once the pattern is identified, the implementation is straightforward.

Maximize Points After Choosing K Tasks Python/Java solution

The implementation sorts tasks by requirement and uses a priority queue to store rewards. Python typically uses heapq with negated values to simulate a max heap, while Java uses PriorityQueue with a reverse comparator. Each iteration selects the largest reward among feasible tasks.

How to solve Maximize Points After Choosing K Tasks in O(n log n)?

First sort all tasks by their requirement. Maintain a pointer that scans the sorted list and a max heap storing rewards of tasks whose requirement is less than or equal to the current points. For each of the k selections, push newly available tasks into the heap and extract the maximum reward. Add it to your current points and continue.

What is the best approach for Maximize Points After Choosing K Tasks?

The most efficient solution uses a greedy strategy combined with sorting and a max heap. Sort tasks by their requirement, push rewards of all currently feasible tasks into a max heap, and repeatedly pick the highest reward task up to k times. This guarantees the maximum gain at each step while keeping future options open. The overall complexity is O(n log n).

Is Maximize Points After Choosing K Tasks asked at Google/Amazon/Meta?

Problems combining greedy selection with a priority queue appear frequently in interviews at companies like Google, Amazon, and Meta. This pattern is similar to well-known interview questions such as IPO or project selection problems where tasks unlock based on current resources.

What data structure is used in Maximize Points After Choosing K Tasks?

The key data structure is a max heap (priority queue). It stores rewards of all tasks that are currently feasible so the algorithm can quickly extract the highest reward task. Sorting the tasks by requirement is also essential to efficiently add newly available tasks.

What is the time complexity of Maximize Points After Choosing K Tasks?

The optimal solution runs in O(n log n) time. Sorting the tasks by requirement costs O(n log n), and each task may be inserted and removed from a priority queue once, which also contributes O(n log n). Space complexity is O(n) for the heap storing available rewards.

Description Hints 3 Solution Video 11:50

3767. Maximize Points After Choosing K Tasks

Medium59.1% acceptance1 company

You are given two integer arrays, technique1 and technique2, each of length n, where n represents the number of tasks to complete.

If the i^th task is completed using technique 1, you earn technique1[i] points.
If it is completed using technique 2, you earn technique2[i] points.

You are also given an integer k, representing the minimum number of tasks that must be completed using technique 1.

You must complete at least k tasks using technique 1 (they do not need to be the first k tasks).

The remaining tasks may be completed using either technique.

Return an integer denoting the maximum total points you can earn.

Example 1:

Input: technique1 = [5,2,10], technique2 = [10,3,8], k = 2

Output: 22

Explanation:

We must complete at least k = 2 tasks using technique1.

Choosing technique1[1] and technique1[2] (completed using technique 1), and technique2[0] (completed using technique 2), yields the maximum points: 2 + 10 + 10 = 22.

Example 2:

Input: technique1 = [10,20,30], technique2 = [5,15,25], k = 2

Output: 60

Explanation:

We must complete at least k = 2 tasks using technique1.

Choosing all tasks using technique 1 yields the maximum points: 10 + 20 + 30 = 60.

Example 3:

Input: technique1 = [1,2,3], technique2 = [4,5,6], k = 0

Output: 15

Explanation:

Since k = 0, we are not required to choose any task using technique1.

Choosing all tasks using technique 2 yields the maximum points: 4 + 5 + 6 = 15.

Constraints:

1 <= n == technique1.length == technique2.length <= 10⁵
1 <= technique1[i], technique2[i] <= 10⁵
0 <= k <= n

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

Initially complete all tasks using way 1; let the initial total be sum(reward1).

Hint 2

The delta when switching task i from way 1 to way 2 is reward2[i] - reward1[i].

Hint 3

Sort deltas descending and apply the largest n - k; after each switch update the maximum.

11:50

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1class Solution:
2    def maxPoints(self, technique1: List[int], technique2: List[int], k: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[5,2,10]
[10,3,8]
2

Input

[10,20,30]
[5,15,25]
2

Input

[1,2,3]
[4,5,6]
0

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