Description#
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.
Implement the PhoneDirectory
class:
PhoneDirectory(int maxNumbers)
Initializes the phone directory with the number of available slots maxNumbers
.int get()
Provides a number that is not assigned to anyone. Returns -1
if no number is available.bool check(int number)
Returns true
if the slot number
is available and false
otherwise.void release(int number)
Recycles or releases the slot number
.
Example 1:
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.
Constraints:
1 <= maxNumbers <= 104
0 <= number < maxNumbers
- At most
2 * 104
calls will be made to get
, check
, and release
.
Solutions#
Solution 1#
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| class PhoneDirectory:
def __init__(self, maxNumbers: int):
"""
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
"""
self.provided = [False] * maxNumbers
def get(self) -> int:
"""
Provide a number which is not assigned to anyone.
@return - Return an available number. Return -1 if none is available.
"""
for i in range(len(self.provided)):
if not self.provided[i]:
self.provided[i] = True
return i
return -1
def check(self, number: int) -> bool:
"""
Check if a number is available or not.
"""
return not self.provided[number]
def release(self, number: int) -> None:
"""
Recycle or release a number.
"""
self.provided[number] = False
# 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)
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| class PhoneDirectory {
private boolean[] provided;
/**
Initialize your data structure here
@param maxNumbers - The maximum numbers that can be stored in the phone directory.
*/
public PhoneDirectory(int maxNumbers) {
provided = 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() {
for (int i = 0; i < provided.length; ++i) {
if (!provided[i]) {
provided[i] = true;
return i;
}
}
return -1;
}
/** Check if a number is available or not. */
public boolean check(int number) {
return !provided[number];
}
/** Recycle or release a number. */
public void release(int number) {
provided[number] = false;
}
}
/**
* 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);
*/
|