Leetcode #1834: Single-Threaded CPU
In this guide, we solve Leetcode #1834 Single-Threaded CPU 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 n tasks labeled from 0 to n - 1 represented by a 2D integer array tasks, where tasks[i] = [enqueueTimei, processingTimei] means that the ith task will be available to process at enqueueTimei and will take processingTimei to finish processing. You have a single-threaded CPU that can process at most one task at a time and will act in the following way: If the CPU is idle and there are no available tasks to process, the CPU remains idle.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Sorting, 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: tasks = [[1,2],[2,4],[3,2],[4,1]]
Output: [0,2,3,1]
Explanation: The events go as follows:
- At time = 1, task 0 is available to process. Available tasks = {0}.
- Also at time = 1, the idle CPU starts processing task 0. Available tasks = {}.
- At time = 2, task 1 is available to process. Available tasks = {1}.
- At time = 3, task 2 is available to process. Available tasks = {1, 2}.
- Also at time = 3, the CPU finishes task 0 and starts processing task 2 as it is the shortest. Available tasks = {1}.
- At time = 4, task 3 is available to process. Available tasks = {1, 3}.
- At time = 5, the CPU finishes task 2 and starts processing task 3 as it is the shortest. Available tasks = {1}.
- At time = 6, the CPU finishes task 3 and starts processing task 1. Available tasks = {}.
- At time = 10, the CPU finishes task 1 and becomes idle.
Python Solution
class Solution:
def getOrder(self, tasks: List[List[int]]) -> List[int]:
for i, task in enumerate(tasks):
task.append(i)
tasks.sort()
ans = []
q = []
n = len(tasks)
i = t = 0
while q or i < n:
if not q:
t = max(t, tasks[i][0])
while i < n and tasks[i][0] <= t:
heappush(q, (tasks[i][1], tasks[i][2]))
i += 1
pt, j = heappop(q)
ans.append(j)
t += pt
return ans
Complexity
The time complexity is , where is the number of tasks. The space complexity is O(n).
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.