LeetCode 379: Design Phone Directory Solution

Master LeetCode problem 379 (Design Phone Directory), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

379. Design Phone Directory

Medium

Array Hash Table Linked List Design Queue

Problem Explanation

Explanation:

To solve this problem, we can use a set to store available phone numbers and a queue to store released phone numbers. When a new phone number is requested, we check if there are any released numbers in the queue. If so, we can assign the released number to the new request. If not, we assign the smallest available number from the set.

When a number is released, we add it to the queue. This way, we can efficiently allocate and deallocate phone numbers in constant time.

: :

Solution Code

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Set;

class PhoneDirectory {
    Set<Integer> availableNumbers;
    Queue<Integer> releasedNumbers;

    public PhoneDirectory(int maxNumbers) {
        availableNumbers = new HashSet<>();
        releasedNumbers = new LinkedList<>();
        for (int i = 0; i < maxNumbers; i++) {
            availableNumbers.add(i);
        }
    }

    public int get() {
        if (availableNumbers.isEmpty()) {
            return -1;
        }
        int num;
        if (!releasedNumbers.isEmpty()) {
            num = releasedNumbers.poll();
        } else {
            num = availableNumbers.iterator().next();
            availableNumbers.remove(num);
        }
        return num;
    }

    public boolean check(int number) {
        return availableNumbers.contains(number) || releasedNumbers.contains(number);
    }

    public void release(int number) {
        if (!check(number)) {
            releasedNumbers.offer(number);
            availableNumbers.add(number);
        }
    }
}

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