all interview problems organized, and cleaned up repeats

src/extra_interview_problems/math_arrays_and_strings/anagram.py

@@ -25,7 +25,24 @@ def is_anagram(s1, s2):
 
     return True
 
+''' verify if words are anagrams by comparing a sum of  Unicode code
+point of the character'''
 
+def get_unicode_sum(word):
+    s = 0
+    for p in word:
+        s += ord(p)
+    return s
+
+
+def is_anagram2(word1, word2):
+    '''
+    >>> is_anagram2('cat', 'tac')
+    True
+    >>> is_anagram2('cat', 'hat')
+    False
+    '''
+    return get_unicode_sum(word1) == get_unicode_sum(word2)
 
 
 if __name__ == '__main__':

src/extra_interview_problems/math_arrays_and_strings/anagrams_hash_table.txt

@@ -1,10 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-Given an english word in the form of a string, how can you quickly find all valid anagrams for that string?
-
-First you go through each word in the dictionary, sort the letters of the word and use this as a key in a Hash Table. O(nln n).
-
-The O(1) lookup is sorting the query word and searching for the key.

src/extra_interview_problems/math_arrays_and_strings/check_if_2_numbers_sum_to_k.py

@@ -17,17 +17,17 @@ Check whether any element of the array appears in the BST.
 It takes O(nlog n) times two.
 """
 
-def check_sum_hash_table(array, k):
+from collections import defaultdict, Counter
+
+def check_sum(array, k):
     '''
-    >>> check_sum_hash_table([3, 2, 6, 7, 9, 1], 8)
+    >>> check_sum([3, 2, 6, 7, 9, 1], 8)
     [(6, 2), (1, 7)]
-    >>> check_sum_hash_table([5, 2, 6, 7, 9, 1], 4)
+    >>> check_sum([5, 2, 6, 7, 9, 1], 4)
     []
     >>>
     '''
 
-    from collections import defaultdict
-
     dict = defaultdict()
     res = []
 
@@ -40,6 +40,32 @@ def check_sum_hash_table(array, k):
 
     return res
 
+
+def check_sum2(array, k):
+    '''
+    >>> check_sum2([1, 4, 2, 7, 1, 3, 10, 15, 3, 1], 6)
+    set([(3, 3)])
+    >>> check_sum2([1, 4, 2, 7, 1, 3, 10, 15, 3, 1], 0)
+    set([])
+    '''
+
+    dict = Counter()
+    res = set()
+
+    for i in array:
+        dict[i] += 1
+
+    for i in array:
+        if dict[k-i] > 0:
+            if i == k-i and dict[k-i] > 1:
+                    res.add((i, k-i))
+                    dict[k-i] -= 2
+            elif i == k-i:
+                res.add((i, k-i))
+                dict[k-i] -= 1
+
+    return res
+
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/extra_interview_problems/math_arrays_and_strings/check_if_anagrams.py

@@ -1,55 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-""" find whether two words are anagrams. Since sets do not count occurrence,
-and sorting is O(nlogn) we will use hash tables. We scan the first string
-and add all the character occurrences. Then we scan the second trying and
-decrease all the character occurrences. If all the counts are zero, it is
-an anagram"""
-
-from collections import Counter
-
-def verify_two_strings_are_anagrams(str1, str2):
-
-    ana_table = Counter()
-
-    for i in str1:
-        ana_table[i] += 1
-
-    for i in str2:
-        ana_table[i] -= 1
-
-    if len(set(ana_table.values())) < 2:
-        return True
-    else:
-        return  False
-
-
-''' verify if words are anagrams by comparing a sum of  Unicode code
-point of the character'''
-
-def get_unicode_sum(word):
-    s = 0
-    for p in word:
-        s += ord(p)
-    return s
-
-
-def find_anagram_get_unicode(word1, word2):
-    return get_unicode_sum(word1) == get_unicode_sum(word2)
-
-
-
-
-if __name__ == '__main__':
-    str1 = 'marina'
-    str2 = 'aniram'
-    str3 = 'anfaam'
-
-    assert(verify_two_strings_are_anagrams(str1, str2) == True)
-    assert(verify_two_strings_are_anagrams(str1, str3) == False)
-
-    assert(find_anagram_get_unicode(str1, str2) == True)
-    assert(find_anagram_get_unicode(str1, str3) == False)

src/extra_interview_problems/math_arrays_and_strings/fibonacci.py

@@ -5,12 +5,38 @@ __author__ = "bt3"
 
 def fib_generator():
     a, b = 0, 1
+
     while True:
         yield b
         a, b = b, a+b
 
 
+def fib(n):
+    '''
+    >>> fib(2)
+    1
+    >>> fib(5)
+    5
+    >>> fib(7)
+    13
+    '''
+    if n < 3:
+        return 1
+
+    a, b = 0, 1
+    count = 1
+
+    while count < n:
+        count += 1
+        a, b = b, a+b
+
+    return b
+
+
 if __name__ == '__main__':
+    import doctest
+    doctest.testmod()
+
     fib = fib_generator()
     print(next(fib))
     print(next(fib))

src/extra_interview_problems/math_arrays_and_strings/find_if_numbers_sum_to_k.py

@@ -1,44 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-'''
-Given an unsorted array of numbers (that may contain repeated numbers),
- * data structure that contains all the pairs w/sum equal to  k.
- * Do not include pairs that are the same numbers in a different order.
-'''
-
-from collections import Counter
-
-def sum_pairs(array, k):
-    '''
-    >>> sum_pairs([1, 4, 2, 7, 1, 3, 10, 15, 3, 1], 6)
-    set([(3, 3)])
-    >>> sum_pairs([1, 4, 2, 7, 1, 3, 10, 15, 3, 1], 0)
-    set([])
-    '''
-
-    results = set()
-    dict = Counter()
-
-    for i in array:
-        dict[i] += 1
-
-    for i in array:
-        if dict[k-i] > 0:
-            if i == k-i and dict[k-i] > 1:
-                    results.add((i, k-i))
-                    dict[k-i] -= 2
-            elif i == k-i:
-                results.add((i, k-i))
-                dict[k-i] -= 1
-
-
-    return results
-
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/extra_interview_problems/math_arrays_and_strings/find_nth_fibonacci.py

@@ -1,38 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-'''
-Write a function to find the nth Fibonacci number.
-Optimize your code so that the numbers
-don't have to be recalculated on consecutive function call
-'''
-
-def fib(n):
-    '''
-    >>> fib(2)
-    1
-    >>> fib(5)
-    5
-    >>> fib(7)
-    13
-    '''
-
-    if n < 3:
-        return 1
-
-    a, b = 0, 1
-    count = 1
-
-    while count < n:
-        count += 1
-        a, b = b, a+b
-
-    return b
-
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/extra_interview_problems/math_arrays_and_strings/finding_if_prime.py

@@ -1,61 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-
-import math
-import random
-
-def finding_prime(number):
-    ''' find whether a number is prime in a simple way'''
-    num = abs(number)
-    if num < 4 : return True
-    for x in range(2, num):
-        if num % x == 0:
-            return False
-    return True
-
-
-def finding_prime_sqrt(number):
-    ''' find whether a number is prime as soon as it rejects all candidates up to sqrt(n) '''
-    num = abs(number)
-    if num < 4 : return True
-    for x in range(2, int(math.sqrt(num)) + 1):
-        if number % x == 0:
-            return False
-    return True
-
-
-def finding_prime_fermat(number):
-    ''' find whether a number is prime with Fermat's theorem, using probabilistic tests '''
-    if number <= 102:
-        for a in range(2, number):
-            if pow(a, number- 1, number) != 1:
-                return False
-        return True
-    else:
-        for i in range(100):
-            a = random.randint(2, number - 1)
-            if pow(a, number - 1, number) != 1:
-                return False
-        return True
-
-
-
-def test_finding_prime():
-    number1 = 17
-    number2 = 20
-    assert(finding_prime(number1) == True)
-    assert(finding_prime(number2) == False)
-    assert(finding_prime_sqrt(number1) == True)
-    assert(finding_prime_sqrt(number2) == False)
-    assert(finding_prime_fermat(number1) == True)
-    assert(finding_prime_fermat(number2) == False)
-    print('Tests passed!')
-
-
-if __name__ == '__main__':
-    test_finding_prime()
-
-

src/extra_interview_problems/math_arrays_and_strings/permutations2.py

@@ -1,29 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-''' Generate all permutations of an alphanumeric string '''
-
-def permutations(word):
-    '''
-    >>> permutations('abc')
-    ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
-    >>> permutations('')
-    ''
-    '''
-    if len(word) < 2:
-        return word
-
-    res = []
-    for i in range(len(word)):
-        rest = word[:i] + word[i+1:]
-        for p in permutations(rest):
-            res.append(word[i] + p)
-    return res
-
-
-
-if __name__ == '__main__':
-    import doctest
-    doctest.testmod()
-

src/extra_interview_problems/math_arrays_and_strings/reverse_words.py

@@ -4,9 +4,9 @@ __author__ = "bt3"
 
 
 
-def invert_word(word):
+def reversing_words(word):
     """
-    >>> invert_word('buffy is awesome')
+    >>> reversing_words('buffy is awesome')
     'awesome is buffy'
     """
     new_word = []
@@ -17,6 +17,28 @@ def invert_word(word):
 
     return " ".join(new_word)
 
+
+def reversing_words2(s):
+    """
+    >>> reversing_words2('buffy is awesome')
+    'awesome is buffy'
+    """
+    words = s.split()
+    return ' '.join(reversed(words))
+
+
+def reversing_words3(s):
+    """
+    >>> reversing_words('buffy is awesome')
+    'awesome is buffy'
+    """
+    words = s.split(' ')
+    words.reverse()
+    return ' '.join(words)
+
+
+
+
 if __name__ == '__main__':
     import doctest
     doctest.testmod()

src/extra_interview_problems/math_arrays_and_strings/reverse_words_sentence.py

@@ -1,24 +0,0 @@
-#!/usr/bin/env python
-
-__author__ = "bt3"
-
-
-def reversing_words_setence_py(s):
-    words = s.split()
-    return ' '.join(reversed(words))
-
-def reversing_words_setence_py2(s):
-    words = s.split(' ')
-    words.reverse()
-    return ' '.join(words)
-
-
-
-
-
-if __name__ == '__main__':
-    s = "Buffy is a Vampire Slayer"
-    print reversing_words_setence_py(s)
-    print reversing_words_setence_py2(s)
-
-