Leetcode #1942: The Number of the Smallest Unoccupied Chair
In this guide, we solve Leetcode #1942 The Number of the Smallest Unoccupied Chair 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
There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: times = [[1,4],[2,3],[4,6]], targetFriend = 1
Output: 1
Explanation:
- Friend 0 arrives at time 1 and sits on chair 0.
- Friend 1 arrives at time 2 and sits on chair 1.
- Friend 1 leaves at time 3 and chair 1 becomes empty.
- Friend 0 leaves at time 4 and chair 0 becomes empty.
- Friend 2 arrives at time 4 and sits on chair 0.
Since friend 1 sat on chair 1, we return 1.
Python Solution
class Solution:
def smallestChair(self, times: List[List[int]], targetFriend: int) -> int:
n = len(times)
for i in range(n):
times[i].append(i)
times.sort()
idle = list(range(n))
heapify(idle)
busy = []
for arrival, leaving, i in times:
while busy and busy[0][0] <= arrival:
heappush(idle, heappop(busy)[1])
j = heappop(idle)
if i == targetFriend:
return j
heappush(busy, (leaving, j))
Complexity
The time complexity is , and the space complexity is . 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.