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February 20, 2013 22:24
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Programming interview practice of the week (Feb. 20, 2013)
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""" | |
Programming interview practice of the week (2013-02-20) | |
Problem of the week - Singletons considered okay | |
1. Write a simple singleton class in your favorite programming language with an example usage. | |
2. Solve the Sorted array binary search problem: | |
Given a sorted array of integers array and an integer key, return the index of the first instance of key in the array. If key is not present in array, you should return -1. | |
For example, given the array [-10, -2, 1, 5, 5, 8, 20] and key 5, you should return the index 3. | |
Your solution should be a binary search, that is, it should run in O(log n) time. | |
""" | |
# Problem 1 | |
class Singleton(object): | |
__instance = None | |
def __new__(cls, *args, **kwargs): | |
if cls.__instance is None: | |
cls.__instance = super(Singleton, cls).__new__(cls, *args, **kwargs) | |
return cls.__instance | |
s = Singleton() | |
t = Singleton() | |
assert s is t | |
# Problem 2 | |
def find(array, key, min_index=0, max_index=None): | |
if max_index is None: | |
max_index = len(array) | |
length = max_index - min_index | |
mid = length / 2 + min_index | |
if length == 0: | |
return -1 | |
elif length == 1: | |
return mid if array[mid] == key else -1 | |
elif array[mid] < key: | |
return find(array, key, mid, max_index) | |
else: | |
lower = find(array, key, min_index, mid) | |
return lower if lower != -1 or array[mid] != key else mid | |
assert find([-10, -2, 1, 5, 5, 8, 20], 5) == 3 |
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