Very Simple & Fast Python Solution using Set

  • 0

    Idea: since phone numbers cannot have duplicates, set is obviously a good option here. check(), release() and get() can be easily done.

    Here is the code:

    class PhoneDirectory(object):
        def __init__(self, maxNumbers):
            # generate number from 0 to maxNumbers
            # and add them into the set
            self.s = set([i for i in range(maxNumbers)])
        def get(self):
            # if set is empty, that means no available number
            if len(self.s) == 0: return -1
            # set.pop() removes an arbitrary element from the set and returns it
            return self.s.pop()
        def check(self, number):
            # check if the number is still available (in the set)
            return number in self.s
        def release(self, number):
            # add the number into the set
            # this is robust since even though number is available already,
            # it won't create a duplicate

    Runtime 115ms. Beats 95.77%

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