Leetcode #1847: Closest Room
In this guide, we solve Leetcode #1847 Closest 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 a hotel with n rooms. The rooms are represented by a 2D integer array rooms where rooms[i] = [roomIdi, sizei] denotes that there is a room with room number roomIdi and size equal to sizei.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Binary Search, Ordered Set, Sorting
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: rooms = [[2,2],[1,2],[3,2]], queries = [[3,1],[3,3],[5,2]]
Output: [3,-1,3]
Explanation: The answers to the queries are as follows:
Query = [3,1]: Room number 3 is the closest as abs(3 - 3) = 0, and its size of 2 is at least 1. The answer is 3.
Query = [3,3]: There are no rooms with a size of at least 3, so the answer is -1.
Query = [5,2]: Room number 3 is the closest as abs(3 - 5) = 2, and its size of 2 is at least 2. The answer is 3.
Python Solution
class Solution:
def closestRoom(
self, rooms: List[List[int]], queries: List[List[int]]
) -> List[int]:
rooms.sort(key=lambda x: x[1])
k = len(queries)
idx = sorted(range(k), key=lambda i: queries[i][1])
ans = [-1] * k
i, n = 0, len(rooms)
sl = SortedList(x[0] for x in rooms)
for j in idx:
prefer, minSize = queries[j]
while i < n and rooms[i][1] < minSize:
sl.remove(rooms[i][0])
i += 1
if i == n:
break
p = sl.bisect_left(prefer)
if p < len(sl):
ans[j] = sl[p]
if p and (ans[j] == -1 or ans[j] - prefer >= prefer - sl[p - 1]):
ans[j] = sl[p - 1]
return ans
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.