379-design-phone-directory
Question
https://leetcode.com/problems/design-phone-directory/description/
Design a Phone Directory which supports the following operations:
- get: Provide a number which is not assigned to anyone.
- check: Check if a number is available or not.
- release: Recycle or release a number.
Example:
// Init a phone directory containing a total of 3 numbers: 0, 1, and 2.
PhoneDirectory directory = new PhoneDirectory(3);
// It can return any available phone number. Here we assume it returns 0.
directory.get();
// Assume it returns 1.
directory.get();
// The number 2 is available, so return true.
directory.check(2);
// It returns 2, the only number that is left.
directory.get();
// The number 2 is no longer available, so return false.
directory.check(2);
// Release number 2 back to the pool.
directory.release(2);
// Number 2 is available again, return true.
directory.check(2);
Thought Process
- Queue and Boolean array
- Time complexity O(1)
- Space complexity O(n)
Solution
class PhoneDirectory {
Queue<Integer> queue;
boolean[] used;
/** Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory. */
public PhoneDirectory(int maxNumbers) {
queue = new LinkedList<>();
for (int i = 0; i < maxNumbers; i++) {
queue.offer(i);
}
used = new boolean[maxNumbers];
}
/** Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available. */
public int get() {
if (queue.isEmpty()) return -1;
used[queue.peek()] = true;
return queue.poll();
}
/** Check if a number is available or not. */
public boolean check(int number) {
return !used[number];
}
/** Recycle or release a number. */
public void release(int number) {
if (used[number]) {
used[number] = false;
queue.offer(number);
}
}
}
/**
* Your PhoneDirectory object will be instantiated and called as such:
* PhoneDirectory obj = new PhoneDirectory(maxNumbers);
* int param_1 = obj.get();
* boolean param_2 = obj.check(number);
* obj.release(number);
*/