Leetcode #1882: Process Tasks Using Servers
In this guide, we solve Leetcode #1882 Process Tasks Using Servers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed integer arrays servers and tasks of lengths n and m respectively. servers[i] is the weight of the ith server, and tasks[j] is the time needed to process the jth task in seconds.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: servers = [3,3,2], tasks = [1,2,3,2,1,2]
Output: [2,2,0,2,1,2]
Explanation: Events in chronological order go as follows:
- At second 0, task 0 is added and processed using server 2 until second 1.
- At second 1, server 2 becomes free. Task 1 is added and processed using server 2 until second 3.
- At second 2, task 2 is added and processed using server 0 until second 5.
- At second 3, server 2 becomes free. Task 3 is added and processed using server 2 until second 5.
- At second 4, task 4 is added and processed using server 1 until second 5.
- At second 5, all servers become free. Task 5 is added and processed using server 2 until second 7.
Python Solution
class Solution:
def assignTasks(self, servers: List[int], tasks: List[int]) -> List[int]:
idle = [(x, i) for i, x in enumerate(servers)]
heapify(idle)
busy = []
ans = []
for j, t in enumerate(tasks):
while busy and busy[0][0] <= j:
_, s, i = heappop(busy)
heappush(idle, (s, i))
if idle:
s, i = heappop(idle)
heappush(busy, (j + t, s, i))
else:
w, s, i = heappop(busy)
heappush(busy, (w + t, s, i))
ans.append(i)
return ans
Complexity
The time complexity is , where is the number of servers and is the number of tasks. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.