Leetcode #855: Exam Room
In this guide, we solve Leetcode #855 Exam Room 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 an exam room with n seats in a single row labeled from 0 to n - 1. When a student enters the room, they must sit in the seat that maximizes the distance to the closest person.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Ordered Set, 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
["ExamRoom", "seat", "seat", "seat", "seat", "leave", "seat"]
[[10], [], [], [], [], [4], []]
Output
[null, 0, 9, 4, 2, null, 5]
Explanation
ExamRoom examRoom = new ExamRoom(10);
examRoom.seat(); // return 0, no one is in the room, then the student sits at seat number 0.
examRoom.seat(); // return 9, the student sits at the last seat number 9.
examRoom.seat(); // return 4, the student sits at the last seat number 4.
examRoom.seat(); // return 2, the student sits at the last seat number 2.
examRoom.leave(4);
examRoom.seat(); // return 5, the student sits at the last seat number 5.
Python Solution
class ExamRoom:
def __init__(self, n: int):
def dist(x):
l, r = x
return r - l - 1 if l == -1 or r == n else (r - l) >> 1
self.n = n
self.ts = SortedList(key=lambda x: (-dist(x), x[0]))
self.left = {}
self.right = {}
self.add((-1, n))
def seat(self) -> int:
s = self.ts[0]
p = (s[0] + s[1]) >> 1
if s[0] == -1:
p = 0
elif s[1] == self.n:
p = self.n - 1
self.delete(s)
self.add((s[0], p))
self.add((p, s[1]))
return p
def leave(self, p: int) -> None:
l, r = self.left[p], self.right[p]
self.delete((l, p))
self.delete((p, r))
self.add((l, r))
def add(self, s):
self.ts.add(s)
self.left[s[1]] = s[0]
self.right[s[0]] = s[1]
def delete(self, s):
self.ts.remove(s)
self.left.pop(s[1])
self.right.pop(s[0])
# Your ExamRoom object will be instantiated and called as such:
# obj = ExamRoom(n)
# param_1 = obj.seat()
# obj.leave(p)
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.