Leetcode #379: Design Phone Directory
In this guide, we solve Leetcode #379 Design Phone Directory 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
Design a phone directory that initially has maxNumbers empty slots that can store numbers. The directory should store numbers, check if a certain slot is empty or not, and empty a given slot.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Design, Queue, Array, Hash Table, Linked List
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
["PhoneDirectory", "get", "get", "check", "get", "check", "release", "check"]
[[3], [], [], [2], [], [2], [2], [2]]
Output
[null, 0, 1, true, 2, false, null, true]
Explanation
PhoneDirectory phoneDirectory = new PhoneDirectory(3);
phoneDirectory.get(); // It can return any available phone number. Here we assume it returns 0.
phoneDirectory.get(); // Assume it returns 1.
phoneDirectory.check(2); // The number 2 is available, so return true.
phoneDirectory.get(); // It returns 2, the only number that is left.
phoneDirectory.check(2); // The number 2 is no longer available, so return false.
phoneDirectory.release(2); // Release number 2 back to the pool.
phoneDirectory.check(2); // Number 2 is available again, return true.
Python Solution
class PhoneDirectory:
def __init__(self, maxNumbers: int):
self.available = set(range(maxNumbers))
def get(self) -> int:
if not self.available:
return -1
return self.available.pop()
def check(self, number: int) -> bool:
return number in self.available
def release(self, number: int) -> None:
self.available.add(number)
# Your PhoneDirectory object will be instantiated and called as such:
# obj = PhoneDirectory(maxNumbers)
# param_1 = obj.get()
# param_2 = obj.check(number)
# obj.release(number)
Complexity
The time complexity is . The space complexity is , where is the value of maxNumbers.
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.