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Leetcode #252: Meeting Rooms

In this guide, we solve Leetcode #252 Meeting Rooms 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.

Leetcode

Problem Statement

Given an array of meeting time intervals where intervals[i] = [starti, endi], determine if a person could attend all meetings. Example 1: Input: intervals = [[0,30],[5,10],[15,20]] Output: false Example 2: Input: intervals = [[7,10],[2,4]] Output: true Constraints: 0 <= intervals.length <= 104 intervals[i].length == 2 0 <= starti < endi <= 106

Quick Facts

  • Difficulty: Easy
  • Premium: Yes
  • Tags: Array, Sorting

Intuition

Sorting reveals structure that is hard to see in the original order.

Once sorted, a linear scan is usually enough to compute the answer.

Approach

Sort the data and sweep through it while maintaining a small state.

This keeps the logic straightforward and reliable.

Steps:

  • Sort the data.
  • Scan in order while maintaining state.
  • Update the best answer.

Example

Input: intervals = [[0,30],[5,10],[15,20]] Output: false

Python Solution

class Solution: def canAttendMeetings(self, intervals: List[List[int]]) -> bool: intervals.sort() return all(a[1] <= b[0] for a, b in pairwise(intervals))

Complexity

The time complexity is O(n×log⁡n)O(n \times \log n)O(n×logn), and the space complexity is O(log⁡n)O(\log n)O(logn), where nnn is the number of meetings. The space complexity is O(log⁡n)O(\log n)O(logn), where nnn is the number of meetings.

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.


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