Design Phone Directory

Design Phone Directory - Problem

You're tasked with designing a phone directory system that manages available phone number slots efficiently. The directory starts with a fixed number of empty slots that can store phone numbers.

Your system needs to support three key operations:

get() - Assign an available phone number slot to a user. Returns the number, or -1 if no slots are available
check(number) - Verify if a specific slot number is currently available
release(number) - Free up a previously assigned slot so it can be reused

Example: If you initialize the directory with 3 slots, numbers 0, 1, and 2 are initially available. After calling get(), one number becomes assigned. You can later release() that number to make it available again.

This problem tests your ability to design efficient data structures for resource management - a common requirement in system design interviews.

Input & Output

example_1.py — Basic Operations

$ Input: ["PhoneDirectory", "get", "check", "get", "check", "release", "check"] [[3], [], [2], [], [2], [2], [2]]

› Output: [null, 0, true, 1, false, null, true]

💡 Note: Initialize with 3 slots (0,1,2). get() returns 0 and marks it unavailable. check(2) returns true as it's still available. get() returns 1. check(2) returns false since 2 is assigned. release(2) frees slot 2. check(2) returns true again.

example_2.py — All Slots Exhausted

$ Input: ["PhoneDirectory", "get", "get", "get", "get"] [[2], [], [], [], []]

› Output: [null, 0, 1, -1, -1]

💡 Note: Initialize with 2 slots. First get() returns 0, second returns 1. Third and fourth get() return -1 since all slots are exhausted.

example_3.py — Release and Reuse

$ Input: ["PhoneDirectory", "get", "get", "release", "get", "release", "get"] [[1], [], [], [0], [], [0], []]

› Output: [null, 0, -1, null, 0, null, -1]

💡 Note: Initialize with 1 slot. get() returns 0. Second get() returns -1. release(0) frees slot 0. get() can now return 0 again. After another release(0) and get(), we're back to no available slots.

Visualization

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Understanding the Visualization

Initialize

All phone numbers 0 to maxNumbers-1 are available

Assignment

get() removes and returns the next available number

Availability Check

check() verifies if a specific number is available

Release

release() returns a number to the available pool

Key Takeaway

🎯 Key Insight: Combining queue for O(1) assignment with set for O(1) lookup creates an optimal resource management system

Time & Space Complexity

Time Complexity

⏱️

O(1)

Queue operations and set lookups are all O(1)

✓ Linear Growth

Space Complexity

O(n)

Queue and set each store up to n numbers

⚡ Linearithmic Space

Constraints

1 ≤ maxNumbers ≤ 10⁴
0 ≤ number < maxNumbers
At most 2 × 10⁴ calls will be made to get, check, and release

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 Apple 15

The optimal solution uses a queue + set combination to achieve O(1) time complexity for all operations. The queue stores available numbers for fast assignment, while the set enables instant availability checking. This approach is highly scalable and commonly used in resource management systems.

Common Approaches

Approach	Time	Space	Notes
✓ Array Linear Search	O(n)	O(n)	Use array to track status, search linearly for available slots
Queue + Set Optimization (Optimal)	O(1)	O(n)	Use queue for O(1) get() and set for O(1) check() operations

Array Linear Search — Algorithm Steps

Initialize boolean array where true means available
For get(): scan array from start to find first true value
For check(): directly access array at given index
For release(): set array[number] = true

Visualization

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Step-by-Step Walkthrough

Initialize

Create boolean array with all slots available (true)

Get Operation

Scan from index 0 until finding first available slot

Mark Assigned

Set found slot to false and return index

Release Operation

Set specified slot back to true

Code -

solution.c — C

typedef struct {
    bool* available;
    int maxNumbers;
} PhoneDirectory;

PhoneDirectory* phoneDirectoryCreate(int maxNumbers) {
    PhoneDirectory* obj = (PhoneDirectory*)malloc(sizeof(PhoneDirectory));
    obj->available = (bool*)malloc(sizeof(bool) * maxNumbers);
    obj->maxNumbers = maxNumbers;
    for (int i = 0; i < maxNumbers; i++) {
        obj->available[i] = true;
    }
    return obj;
}

int phoneDirectoryGet(PhoneDirectory* obj) {
    for (int i = 0; i < obj->maxNumbers; i++) {
        if (obj->available[i]) {
            obj->available[i] = false;
            return i;
        }
    }
    return -1;
}

bool phoneDirectoryCheck(PhoneDirectory* obj, int number) {
    if (number >= 0 && number < obj->maxNumbers) {
        return obj->available[number];
    }
    return false;
}

void phoneDirectoryRelease(PhoneDirectory* obj, int number) {
    if (number >= 0 && number < obj->maxNumbers) {
        obj->available[number] = true;
    }
}

void phoneDirectoryFree(PhoneDirectory* obj) {
    free(obj->available);
    free(obj);
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

get() requires linear scan through array in worst case

✓ Linear Growth

Space Complexity

O(n)

Boolean array to store availability status of n slots

⚡ Linearithmic Space

Constraints

1 ≤ maxNumbers ≤ 10⁴
0 ≤ number < maxNumbers
At most 2 × 10⁴ calls will be made to get, check, and release

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