Practice/Apple/Leetcode 379. Design Phone Directory

Leetcode 379. Design Phone Directory

CodingMust

Problem

Design and implement a resource pool manager that handles a fixed collection of numbered resources (from 0 to maxResources-1). Your implementation should efficiently support the following operations:

• Allocate: Assign and return an available resource identifier
• Check: Determine whether a specific resource is currently available
• Release: Free a previously allocated resource, returning it to the available pool

The system must track which resources are currently in use and which are available for allocation. When a resource is released, it should become available for future allocation requests.

Requirements

1. Implement a constructor that initializes the pool with a specified maximum number of resources
2. The get() method should return an available resource ID, or -1 if none are available
3. The check(resourceId) method should return true if the resource is available, false otherwise
4. The release(resourceId) method should mark a resource as available for future use
5. All operations should ideally run in O(1) time complexity
6. Handle edge cases like releasing already-available resources or invalid resource IDs gracefully

Constraints

• 1 ≤ maxResources ≤ 10^4
• 0 ≤ resourceId < maxResources
• At most 2 × 10^4 total calls will be made to get, check, and release
• Resource IDs are integers in the range [0, maxResources-1]

Examples

Example 1:

` Operations: Initialize(3), get(), get(), check(0), release(0), check(0), get() Output: [null, 0, 1, false, null, true, 0]

Explanation:

• ResourcePool(3): Initialize with resources 0, 1, 2
• get(): Returns 0 (first available resource)
• get(): Returns 1 (next available resource)
• check(0): Returns false (resource 0 is allocated)
• release(0): Release resource 0 back to the pool
• check(0): Returns true (resource 0 is now available)
• get(): Returns 0 (resource 0 is available again) `

Example 2:

` Operations: Initialize(2), get(), get(), get() Output: [null, 0, 1, -1]

Explanation:

• ResourcePool(2): Initialize with resources 0, 1
• get(): Returns 0
• get(): Returns 1
• get(): Returns -1 (no more resources available) `

Example 3:

` Operations: Initialize(5), get(), get(), release(0), get(), check(0) Output: [null, 0, 1, null, 0, false]

Explanation:

• After releasing resource 0, it becomes available for reallocation
• The next get() returns resource 0 again
• Checking resource 0 shows it's now allocated (unavailable) `