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Leetcode #1229: Meeting Scheduler

In this guide, we solve Leetcode #1229 Meeting Scheduler 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 the availability time slots arrays slots1 and slots2 of two people and a meeting duration duration, return the earliest time slot that works for both of them and is of duration duration. If there is no common time slot that satisfies the requirements, return an empty array.

Quick Facts

  • Difficulty: Medium
  • Premium: Yes
  • Tags: Array, Two Pointers, Sorting

Intuition

The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.

Moving one pointer at a time keeps the invariant intact and avoids nested loops.

Approach

Place pointers at the left and right ends and move them based on the comparison or target condition.

This yields a clean linear pass after any required sorting.

Steps:

  • Set left and right pointers.
  • Move a pointer based on the condition.
  • Update the best answer while scanning.

Example

Input: slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 8 Output: [60,68]

Python Solution

class Solution: def minAvailableDuration( self, slots1: List[List[int]], slots2: List[List[int]], duration: int ) -> List[int]: slots1.sort() slots2.sort() m, n = len(slots1), len(slots2) i = j = 0 while i < m and j < n: start = max(slots1[i][0], slots2[j][0]) end = min(slots1[i][1], slots2[j][1]) if end - start >= duration: return [start, start + duration] if slots1[i][1] < slots2[j][1]: i += 1 else: j += 1 return []

Complexity

The time complexity is O(m×log⁡m+n×log⁡n)O(m \times \log m + n \times \log n)O(m×logm+n×logn), and the space complexity is O(log⁡m+log⁡n)O(\log m + \log n)O(logm+logn). The space complexity is O(log⁡m+log⁡n)O(\log m + \log n)O(logm+logn).

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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