diff --git a/src/extra_interview_problems/bit/bitwise.txt b/src/extra_interview_problems/bitwise/bitwise.txt
similarity index 100%
rename from src/extra_interview_problems/bit/bitwise.txt
rename to src/extra_interview_problems/bitwise/bitwise.txt
diff --git a/src/extra_interview_problems/bitwise/clear_bits.py b/src/extra_interview_problems/bitwise/clear_bits.py
new file mode 100755
index 0000000..36d477b
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/clear_bits.py
@@ -0,0 +1,35 @@
+#!/usr/bin/python
+
+''' Clear a bit in a binary number.
+    Like the reverse of set bit:
+    1) first create a number filled of 1s,
+        with 0 at i (can create 0001000 and ~)
+    2) AND the number so it clears the ith bit
+'''
+
+
+
+def clear_bit(num, i):
+    mask = ~ (1 << i)   # -0b10001
+    return bin(num & mask)
+
+
+def clear_all_bits_from_i_to_0(num, i):
+    mask = ~ ( (1 << (i+1)) - 1)
+    return bin(num & mask)
+
+
+def clear_all_bits_from_most_sig_to_1(num, i):
+    mask =  ( 1 << i) -1
+    return bin(num & mask)
+
+
+if __name__ == '__main__':
+    num = int('10010000', 2)
+    print clear_bit(num, 4)   # '0b10000000'
+
+    num = int('10010011', 2)
+    print clear_all_bits_from_i_to_0(num, 2)  # '0b10010000'
+
+    num = int('1110011', 2)
+    print clear_all_bits_from_most_sig_to_1(num, 2)  #'0b11'
diff --git a/src/extra_interview_problems/bitwise/find_bit_len.py b/src/extra_interview_problems/bitwise/find_bit_len.py
new file mode 100755
index 0000000..03812ab
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/find_bit_len.py
@@ -0,0 +1,23 @@
+#!/usr/bin/python
+
+''' Find how many bits a int has:
+    1) Start with a mask of 1
+    2) Mask with AND
+    3) if result (if true): count += 1
+    (obs: to find the int of a bin do int('1001', 2)) and to show in bin do bin(int))
+'''
+
+
+def find_bit_len(int_num):
+    lenght = 0
+    while int_num:
+        int_num >>= 1
+        lenght += 1
+    return lenght
+
+
+if __name__ == '__main__':
+    for i in range(17):
+        print(find_bit_len(i))
+        print i.bit_length()
+        print
diff --git a/src/extra_interview_problems/bitwise/get_bit.py b/src/extra_interview_problems/bitwise/get_bit.py
new file mode 100755
index 0000000..994734f
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/get_bit.py
@@ -0,0 +1,25 @@
+#!/usr/bin/python
+
+''' Get a bit in a binary number:
+    1) Shifts 1 over by i bits
+    2) make an AND with the number
+    3) all the other than the bit at i are clean, now compare to 0
+    4) if the new value is not 0, bit i is 1
+'''
+
+
+def get_bit(num, i):
+    mask = 1 << i
+    return num & mask != 0
+
+
+if __name__ == '__main__':
+    num = int('0100100', 2)
+    get_bit(num, 0)   # 0
+    get_bit(num, 1)   # 0
+    get_bit(num, 2)   # 1
+    get_bit(num, 3)   # 0
+    get_bit(num, 4)   # 0
+    get_bit(num, 5)   # 1
+    get_bit(num, 6)   # 0
+
diff --git a/src/extra_interview_problems/bitwise/num_bits_to_convert_2_nums.py b/src/extra_interview_problems/bitwise/num_bits_to_convert_2_nums.py
new file mode 100755
index 0000000..236f00f
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/num_bits_to_convert_2_nums.py
@@ -0,0 +1,37 @@
+#!/usr/bin/python
+
+''' This method returns the number of bits that are necessary to change to convert two
+    numbers A and B:
+    1) XOR
+    2) count 1s
+'''
+
+def count_bits_swap2(a, b):
+    count = 0
+    m = a^b
+    while m:
+        count +=1
+        m = m & (m-1)
+    return count
+
+
+
+def count_bits_swap(a, b):
+    m = a^b
+    return count_1s(m)
+
+
+def count_1s(m):
+    count = 0
+    while m:
+        if m& 1 :
+            count +=1
+        m >>= 1
+    return count
+
+
+if __name__ == '__main__':
+    a = int('10010000', 2)
+    b = int('01011010', 2)
+    print count_bits_swap(a, b)  #4
+    print count_bits_swap2(a, b)  #4
diff --git a/src/extra_interview_problems/bitwise/set_bit.py b/src/extra_interview_problems/bitwise/set_bit.py
new file mode 100755
index 0000000..53896c1
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/set_bit.py
@@ -0,0 +1,25 @@
+#!/usr/bin/python
+''' Set a bit in a binary number:
+    1) Shifts 1 over by i bits
+    2) make an OR with the number, only the value at bit i will change and all the others bit
+    of the mask are zero so will not affect the num
+
+'''
+
+
+
+def set_bit(num, i):
+    mask = 1 << i
+    return bin( num | mask )
+
+
+
+
+if __name__ == '__main__':
+    num = int('0100100', 2)
+    print set_bit(num, 0)   #'0b100101'
+    print set_bit(num, 1)   #'0b100110'
+    print set_bit(num, 2)   # nothing change '0b100100'
+    print set_bit(num, 3)   #'0b101100'
+    print set_bit(num, 4) #'0b110100'
+    print set_bit(num, 5)     # nothing change '0b100100'
diff --git a/src/extra_interview_problems/bitwise/update_bit.py b/src/extra_interview_problems/bitwise/update_bit.py
new file mode 100755
index 0000000..41d3bd6
--- /dev/null
+++ b/src/extra_interview_problems/bitwise/update_bit.py
@@ -0,0 +1,20 @@
+#!/usr/bin/python
+
+''' This method merges set bit and clean bit:
+    1) first clear the bit at i using a mask such as 1110111
+    2) then shift the intended value v by i bits
+    3) this will create a number with bit i to v and all other to 0
+    4) finally update the ith bit with or
+'''
+
+
+
+def  update_bit(num, i, v):
+    mask = ~ (1 << i)
+    return bin( (num & mask) | (v << i) )
+
+
+if __name__ == '__main__':
+    num = int('10010000', 2)
+    print update_bit(num, 2, 1)   # '0b10010100'
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/anagrams_hash_table.txt b/src/extra_interview_problems/math_arrays_and_strings/anagrams_hash_table.txt
new file mode 100755
index 0000000..7f1f14c
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/anagrams_hash_table.txt
@@ -0,0 +1,5 @@
+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.
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/balance.txt b/src/extra_interview_problems/math_arrays_and_strings/balance.txt
new file mode 100755
index 0000000..6f081f1
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/balance.txt
@@ -0,0 +1,6 @@
+This is the classic "you have 8 balls/coins, which are the same weight, except for one which is slightly heavier than the others. You also have an old-style balance. What is the fewest number of weighings to find the heavy coin/ball?
+
+Answer: 2! You need to use every information available:
+Weight 3 x 3 balls/coins.
+If they weight the same: weight the 2 balls/coins left outside.
+Else, measure 2 of the 3 heavier balls/coins.
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/check_2_numbers_array_sum.py b/src/extra_interview_problems/math_arrays_and_strings/check_2_numbers_array_sum.py
new file mode 100755
index 0000000..2a9f7df
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/check_2_numbers_array_sum.py
@@ -0,0 +1,54 @@
+#!/bin/python
+
+
+"""
+Given an integer x and an unsorted array of integers, describe an algorithm to
+determine whether two of the numbers add up to x.
+
+1. Using hash tables.
+2. Sorting the array and keeping two pointers in the array, one in the beginning and
+one in the end. Whenever the sum of the current two integers is less than x, move the
+first pointer forwards, and whenever the sum is greater than x, move the second pointer
+backwards. O(nln n).
+3. Create a BST with x minus each element in the array. Check whether any element of
+the array appears in the BST. It takes O(nlog n) times two.
+"""
+
+
+def check_if_sum(arr, num):
+    seq =  sorted(arr)
+    p1, p2 = 0, len(arr) - 1
+    while p1 < p2 and p2 < len(arr) and p1 >= 0:
+        sum_here = seq[p1] + seq[p2]
+        if sum_here == num:
+            return True, seq[p1], seq[p2]
+        elif sum_here < num:
+            p1 += 1
+        elif sum_here > num:
+            p2 -= 1
+    return False
+
+
+from collections import Counter
+def check_if_sum2(arr, num):
+    d = Counter()
+    for i in arr:
+        d[i] += 1
+
+    for i in arr:
+        other = num - i
+        d[i] -= 1
+        if d[other] == 1:
+            return True, other, i
+    return False
+
+
+arr = [3, 1, 13, 7, 2, 10]
+num1 = 11
+num2 = 6
+
+print(check_if_sum(arr, num1))
+print(check_if_sum(arr, num2))
+print
+print(check_if_sum2(arr, num1))
+print(check_if_sum2(arr, num2))
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/check_if_anagrams.py b/src/extra_interview_problems/math_arrays_and_strings/check_if_anagrams.py
new file mode 100755
index 0000000..b2cf38f
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/check_if_anagrams.py
@@ -0,0 +1,53 @@
+#!/usr/bin/python
+
+""" find whether two words are anagrams. Since sets do not count occurency, and sorting is O(nlogn)
+we will use hash tables. We scan the first string and add all the character occurences. Then we
+scan the second tring and decrease all the caracther occurences. If all the counts are zero, it is
+an anagram"""
+
+import string
+
+
+def verify_two_strings_are_anagrams(str1, str2):
+    ana_table = {key:0 for key in string.ascii_lowercase}
+
+    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 comparying hash functions'''
+
+def hash_func(astring, tablesize):
+    sump = 0
+    for p in astring:
+        sump += ord(p)
+    return sump%tablesize
+
+
+def find_anagram_hash_function(word1, word2):
+    tablesize = 11
+    return hash_func(word1, tablesize) == hash_func(word2, tablesize)
+
+
+
+
+if __name__ == '__main__':
+    str1 = 'marina'
+    str2 = 'aniram'
+    str3 = 'anfaam'
+
+    print verify_two_strings_are_anagrams(str1, str2)
+    print verify_two_strings_are_anagrams(str1, str3)
+    print
+    print find_anagram_hash_function(str1, str2)
+    print find_anagram_hash_function(str1, str3)
+
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/check_palindrome.py b/src/extra_interview_problems/math_arrays_and_strings/check_palindrome.py
new file mode 100644
index 0000000..d17116b
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/check_palindrome.py
@@ -0,0 +1,38 @@
+#!/bin/python
+
+''' Check if palindrome'''
+
+
+def check_pal(string):
+    string = "".join(string.split(' '))
+    p1, p2 = 0, len(string)-1
+    pal = True
+    while p1 < p2:
+        if string[p1] != string[p2]:
+            pal = False
+            break
+        p1+=1
+        p2-=1
+    return pal
+
+
+def check_pal2(string):
+    string = "".join(string.split(' '))
+    if len(string)<2:
+        return True
+    if string[0] != string[-1]:
+        return False
+    return check_pal2(string[1:-1])
+
+
+if __name__ == '__main__':
+    string1 = "borrow or rob"
+    string2 = " draw ward"
+    string3 = "google is fun"
+    print(check_pal(string1))
+    print(check_pal(string2))
+    print(check_pal(string3))
+    print
+    print(check_pal2(string1))
+    print(check_pal2(string2))
+    print(check_pal2(string3))
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/delete_duplicate_char_str.py b/src/extra_interview_problems/math_arrays_and_strings/delete_duplicate_char_str.py
new file mode 100755
index 0000000..5f962ae
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/delete_duplicate_char_str.py
@@ -0,0 +1,23 @@
+#!/usr/bin/python
+
+'''  find and delete all the duplicate characters in a string '''
+
+
+import string
+
+def delete_unique_word(str1):
+    table_c = { key : 0  for key in string.ascii_lowercase}
+    for i in str1:
+        table_c[i] += 1
+    for key, value in table_c.items():
+        if value > 1:
+            str1 = str1.replace(key, "")
+
+    return str1
+
+
+
+if __name__ == '__main__':
+    str1 = "google"
+    print delete_unique_word(str1) # == 'le'
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/fibonacci.py b/src/extra_interview_problems/math_arrays_and_strings/fibonacci.py
new file mode 100755
index 0000000..28ca28b
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/fibonacci.py
@@ -0,0 +1,21 @@
+#!/usr/bin/env python
+
+
+__author__ = "Mari Wahl"
+__email__ = "marina.w4hl@gmail.com"
+
+ ''' understanding generators'''
+
+def fib_generator():
+    a, b = 0, 1
+    while True:
+        yield b
+        a, b = b, a+b
+
+
+if __name__ == '__main__':
+    fib = fib_generator()
+    print(next(fib))
+    print(next(fib))
+    print(next(fib))
+    print(next(fib))
diff --git a/src/extra_interview_problems/math_arrays_and_strings/find_all_permutations_string.py b/src/extra_interview_problems/math_arrays_and_strings/find_all_permutations_string.py
new file mode 100755
index 0000000..8d80954
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/find_all_permutations_string.py
@@ -0,0 +1,81 @@
+#!/usr/bin/python
+
+
+def find_permutations(s):
+    if len(s) < 2: return s
+    res = []
+    for i, c in enumerate(s):
+        for perm in find_permutations(s[:i] + s[i+1:]):
+            res.append(c + "".join(perm))
+    return res
+
+
+def fpc(s):
+    return [s] if len(s)<2 else [c+p for i,c in enumerate(s) for p in fpc(s[:i]+s[i+1:])]
+
+
+def find_permutations_stdlib(s):
+    from itertools import permutations
+    return [''.join(p) for p in permutations(s)]
+
+
+def find_permutations2(s):
+    if len(s) < 2: return s
+    res = []
+    for i in range(len(s)):
+        for perm in find_permutations2(s[:i] + s[i+1:]):
+            res.append(s[i] + perm)
+    return res
+
+
+def verify_if_perm(s1, s2):
+    if len(s1) != len(s2): return False
+    s1 = sorted(s1)
+    s2 = sorted(s2)
+    return s1 == s2
+
+
+from collections import Counter
+def verify_if_perm2(s1, s2): # like anagram
+    if len(s1) != len(s2): return False
+    dict_aux = Counter()
+    for c in s1:
+        dict_aux[c] += 1
+    for c in s2:
+        dict_aux[c] -= 1
+    for item in dict_aux:
+        if dict_aux[item]:
+            return False
+    return True
+
+
+
+if __name__ == '__main__':
+    s1 = 'ufyfbufyfb'
+    s2 = 'buffybuffy'
+    s3 = 'uuyfbuuyfb'
+    s4 = ''
+
+    print(verify_if_perm(s1, s2))
+    print(verify_if_perm(s1, s3))
+    print(verify_if_perm(s1, s4))
+    print
+    print(verify_if_perm2(s1, s2))
+    print(verify_if_perm2(s1, s3))
+    print(verify_if_perm2(s1, s4))
+
+    print
+
+    s = 'hat'
+    print find_permutations(s)
+    print fpc(s)
+    print find_permutations_stdlib(s)
+    print find_permutations2(s)
+    print
+    print find_permutations(s4)
+    print fpc(s4)
+    print find_permutations_stdlib(s4)
+    print find_permutations2(s4)
+
+
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/find_combinations_strings.py b/src/extra_interview_problems/math_arrays_and_strings/find_combinations_strings.py
new file mode 100755
index 0000000..c0ea950
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/find_combinations_strings.py
@@ -0,0 +1,20 @@
+#!/usr/bin/python
+#
+''' give all the combinations of a str or list '''
+
+
+def comb_str(l1):
+    if len(l1) < 2:
+        return l1
+    result = []
+    for i, c in enumerate(l1):
+        result.append(c)
+        for comb in comb_str(l1[i+1:]):
+            result.append(c + "".join(comb))
+    return result
+
+
+
+if __name__ == '__main__':
+    l1 = ['a', 'b', 'c']
+    print comb_str(l1)
diff --git a/src/extra_interview_problems/math_arrays_and_strings/find_if_is_substr.py b/src/extra_interview_problems/math_arrays_and_strings/find_if_is_substr.py
new file mode 100755
index 0000000..3fcc0a7
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/find_if_is_substr.py
@@ -0,0 +1,31 @@
+#!/usr/bin/python
+
+def isSubstr(s1, s2):
+    if s1 in s2 or s2 in s1: return True
+    return False
+
+
+def find_substr(s1, s2):
+    pl, ps = 0, 0
+    if len(s1) > len(s2):
+        larger, smaller = s1, s2
+    else:
+        larger, smaller = s2, s1
+    while ps < len(smaller) and pl < len(larger):
+        if larger[pl] == smaller[ps]:
+            ps += 1
+        else:
+            ps = 0
+        pl += 1
+        if ps == len(smaller):
+            return True
+    return False
+
+
+
+if __name__ == '__main__':
+    s1 = 'buffy is a vampire slayer'
+    s2 = 'vampire'
+    s3 = 'angel'
+    print find_substr(s2, s1)
+    print find_substr(s3, s1)
diff --git a/src/extra_interview_problems/math_arrays_and_strings/find_non_repeating_number.py b/src/extra_interview_problems/math_arrays_and_strings/find_non_repeating_number.py
new file mode 100755
index 0000000..b440014
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/find_non_repeating_number.py
@@ -0,0 +1,45 @@
+#!/usr/bin/python
+
+from collections import defaultdict
+
+
+def find_unique_number(arr):
+    table = defaultdict(int)
+    total = 0
+    for i in arr:
+        if table[i]:
+            total -= i
+        else:
+            total += i
+            table[i] +=1
+    return total
+
+
+def find_unique_number_xor(arr):
+    xor = 0
+    for item in arr:
+        xor ^= item
+    return xor
+
+
+def find_unique_char(s):
+    if len(s) < 2: return True
+    for i, c in enumerate(s):
+        for j in s[i+1:]:
+            if j == c:
+                return False
+    return True
+
+
+
+if __name__ == '__main__':
+    arr = [1, 3, 5, 6, 1, 5, 6, 3, 7,]
+    print(find_unique_number(arr))
+    print(find_unique_number_xor(arr))
+    s1 = 'abcdefg'
+    s2 = 'buffy'
+    s3 = ''
+    print
+    print(find_unique_char(s1))
+    print(find_unique_char(s2))
+    print(find_unique_char(s3))
diff --git a/src/extra_interview_problems/math_arrays_and_strings/finding_gcd.py b/src/extra_interview_problems/math_arrays_and_strings/finding_gcd.py
new file mode 100755
index 0000000..12163d7
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/finding_gcd.py
@@ -0,0 +1,26 @@
+#!/usr/bin/python
+
+
+def finding_gcd(a, b):
+    ''' implements the greatest common divider algorithm '''
+    while(b != 0):
+        result = b
+        a, b = b, a % b        
+    return result
+
+
+def test_finding_gcd():
+    number1 = 21
+    number2 = 12
+    assert(finding_gcd(number1, number2) == 3)
+    print('Tests passed!')
+
+if __name__ == '__main__':
+    test_finding_gcd()
+
+
+
+    
+    
+    
+    
diff --git a/src/extra_interview_problems/math_arrays_and_strings/finding_if_prime.py b/src/extra_interview_problems/math_arrays_and_strings/finding_if_prime.py
new file mode 100755
index 0000000..2c2c447
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/finding_if_prime.py
@@ -0,0 +1,59 @@
+#!/usr/bin/python3
+# mari von steinkirch @2013
+# steinkirch at gmail
+
+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()
+
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/generate_prime.py b/src/extra_interview_problems/math_arrays_and_strings/generate_prime.py
new file mode 100755
index 0000000..a008b4c
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/generate_prime.py
@@ -0,0 +1,35 @@
+#!/usr/bin/python3
+# mari von steinkirch @2013
+# steinkirch at gmail
+
+import math
+import random
+import sys
+from finding_prime import finding_prime_sqrt
+
+
+def generate_prime(number=3):
+    ''' return a n-bit prime '''
+    while 1:
+        p = random.randint(pow(2, number-2), pow(2, number-1)-1)
+        p = 2 * p + 1
+        if finding_prime_sqrt(p):
+            return p
+
+
+if __name__ == '__main__':
+    if len(sys.argv) < 2:
+        print ("Usage: generate_prime.py number")
+        sys.exit()
+    else:
+        number = int(sys.argv[1])
+        print(generate_prime(number))
+
+
+
+
+
+
+
+
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/get_float_rep_bin.py b/src/extra_interview_problems/math_arrays_and_strings/get_float_rep_bin.py
new file mode 100755
index 0000000..5c62268
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/get_float_rep_bin.py
@@ -0,0 +1,26 @@
+#!/usr/bin/python
+
+''' Given a real number between 0 and 1 (eg: 0.72), this method print the binary
+    representation. If the Number cannot be represented accurately in binary, with at
+    most 32 chars, print error:
+'''
+
+def get_float_rep(num):
+    if num >= 1 or num <= 0: return 'Error 1'
+    result = '.'
+    while num:
+        if len(result) >= 32: return 'Error 2', result
+        r = num*2
+        if r >= 1:
+            result += '1'
+            num = r - 1
+        else:
+            result += '0'
+            num = r
+    return result
+
+
+if __name__ == '__main__':
+    print get_float_rep(0.72) #('Error 2', '.1011100001010001111010111000010')
+    print get_float_rep(0.1) # ('Error 2', '.0001100110011001100110011001100')
+    print get_float_rep(0.5) #'.1'
diff --git a/src/extra_interview_problems/math_arrays_and_strings/interserction_two_arrays.py b/src/extra_interview_problems/math_arrays_and_strings/interserction_two_arrays.py
new file mode 100755
index 0000000..485634a
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/interserction_two_arrays.py
@@ -0,0 +1,80 @@
+#!/usr/bin/python
+
+__author__ = "Mari Wahl"
+__email__ = "marina.w4hl@gmail.com"
+
+
+
+''' using sets '''
+
+def intersection_two_arrays_sets(seq1, seq2):
+    ''' find the intersection of two arrays using set proprieties '''
+    set1 = set(seq1)
+    set2 = set(seq2)
+    return set1.intersection(set2)  #same as list(set1 & set2
+
+
+
+''' using merge sort '''
+
+def intersection_two_arrays_ms(seq1, seq2):
+    ''' find the intersection of two arrays using merge sort '''
+    res = []
+    while seq1 and seq2:
+        if seq1[-1] == seq2[-1]:
+            res.append(seq1.pop())
+            seq2.pop()
+        elif seq1[-1] > seq2[-1]:
+            seq1.pop()
+        else:
+            seq2.pop()
+    res.reverse()
+    return res
+
+
+
+
+''' using binary search '''
+
+def binary_search(seq, key, lo=0, hi=None):
+    ''' binary search iterative algorithm '''
+    hi = hi or len(seq)
+    while lo < hi:
+        mid = (hi+lo) // 2
+        if seq[mid] == key:
+            return True
+        elif key > seq[mid]:
+            lo = mid + 1
+        else:
+            hi = mid
+    return None
+
+def intersection_two_arrays_bs(seq1, seq2):
+    ''' if A small and B is too large, we can do a binary search on each entry in B '''
+    ''' only works if sorted and the small sequence has not larger nmbers!!!'''
+    if len(seq1) > len(seq2): seq, key = seq1, seq2
+    else: seq, key = seq2, seq1
+
+    intersec = []
+    for item in key:
+        if binary_search(seq, item):
+            intersec.append(item)
+    return intersec
+
+
+
+def test_intersection_two_arrays(module_name='this module'):
+    seq1 = [1,2,3,5,7,8]
+    seq2 = [3,5,6]
+
+    assert(set(intersection_two_arrays_sets(seq1,seq2)) == set([3,5]))
+    assert(intersection_two_arrays_bs(seq1,seq2) == [3,5])
+    assert(intersection_two_arrays_ms(seq1,seq2) == [3,5])
+
+    s = 'Tests in {name} have {con}!'
+    print(s.format(name=module_name, con='passed'))
+
+
+if __name__ == '__main__':
+    test_intersection_two_arrays()
+
diff --git a/src/extra_interview_problems/math_arrays_and_strings/max_subarray_stocks.py b/src/extra_interview_problems/math_arrays_and_strings/max_subarray_stocks.py
new file mode 100755
index 0000000..b3acaf9
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/max_subarray_stocks.py
@@ -0,0 +1,31 @@
+def beating_stock(array):
+
+    imin = 0
+    i = 1
+    deal = [array[i] - array[imin], imin, i]
+
+    while i < len(array):
+
+        deal_here = array[i] - array[imin]
+        if deal_here > deal[0]:
+            deal = [deal_here, imin, i]
+
+        elif array[i] < array[imin]:
+            imin = i
+
+        i += 1
+
+    return deal[0], array[deal[1]], array[deal[2]]
+
+
+
+
+
+
+
+
+
+
+array = [7, 2, 3, 6, 5, 8, 5, 3, 4]
+print(array)
+print("The best profit is %d, buying at %d, selling at %d." %(beating_stock(array)))
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/number_of_zeros_factorial.txt b/src/extra_interview_problems/math_arrays_and_strings/number_of_zeros_factorial.txt
new file mode 100755
index 0000000..317bfa1
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/number_of_zeros_factorial.txt
@@ -0,0 +1,9 @@
+How to know how many 0 are in 100!
+
+You look for the primes that multiply to 10, i.e. 2 and 5.
+
+There are more 5 than 2s so you can count the fives.
+
+there is 100/5 = 20, so 20 5s. However, there are two 5s in 25, 50, 75 and 100.
+
+result: 20+4 = 24
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/ransom_note.py b/src/extra_interview_problems/math_arrays_and_strings/ransom_note.py
new file mode 100755
index 0000000..c81a680
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/ransom_note.py
@@ -0,0 +1,28 @@
+from collections import Counter
+
+def check_if_ransom_note(magazines, note):
+    count = Counter()
+    pm, pn = 0, 0
+
+    while pn < len(note) and pm < len(magazines):
+        char_note = note[pn]
+        if count[char_note]>0:
+            count[char_note] -= 1
+            pn += 1
+        else:
+            char_magazine = magazines[pm]
+            count[char_magazine] += 1
+            pm +=1
+
+    return pn == len(note)
+
+
+
+
+
+magazines1 = "avfegthhgrebvkdsvnijnvyijfdmckdsmovkmmfvskumvl;cdkmioswckofjbkreenyukjemjgnmkmvkmnvdkmvkr g gmvdvmldm vldfkmbldkmlvdkm"
+magazines2 = "adfsfa"
+note = "you should disobey"
+
+print(check_if_ransom_note(magazines1, note))
+print(check_if_ransom_note(magazines2, note))
diff --git a/src/extra_interview_problems/math_arrays_and_strings/rev_string.py b/src/extra_interview_problems/math_arrays_and_strings/rev_string.py
new file mode 100644
index 0000000..c3da5be
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/rev_string.py
@@ -0,0 +1,12 @@
+#!/bin/python
+
+''' Reverting a string '''
+
+
+def revert(string):
+    return string[::-1]
+
+
+if __name__ == '__main__':
+    string = "Google is fun!"
+    print(revert(string))
\ No newline at end of file
diff --git a/src/extra_interview_problems/math_arrays_and_strings/reverse_words_sentence.py b/src/extra_interview_problems/math_arrays_and_strings/reverse_words_sentence.py
new file mode 100755
index 0000000..5904109
--- /dev/null
+++ b/src/extra_interview_problems/math_arrays_and_strings/reverse_words_sentence.py
@@ -0,0 +1,43 @@
+#!/usr/bin python
+
+
+
+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)
+
+def reversing_words_setence_py3(s):
+    p1 = 0
+    word = ''
+    arr = []
+    while p1 < len(s):
+        if s[p1] != ' ':
+            word += s[p1]
+        else:
+            arr.append(word)
+            word = ''
+        p1 += 1
+
+    arr.append(word)
+    new = ''
+    while arr:
+        new += arr.pop()
+        new += ' '
+
+    return new
+
+
+
+
+if __name__ == '__main__':
+    s = "Buffy is a Vampire Slayer"
+    print reversing_words_setence_py(s)
+    print reversing_words_setence_py2(s)
+    print reversing_words_setence_py3(s)
+
+
diff --git a/src/extra_interview_problems/sorting_and_searching/deck_shuffling.txt b/src/extra_interview_problems/sorting_and_searching/deck_shuffling.txt
new file mode 100755
index 0000000..2f9e063
--- /dev/null
+++ b/src/extra_interview_problems/sorting_and_searching/deck_shuffling.txt
@@ -0,0 +1,3 @@
+How to shuffle an array or a deck of cards so each shuffle has the same probability?
+
+go thru the array shuffling each element with any random element after that.
\ No newline at end of file
diff --git a/src/extra_interview_problems/sorting_and_searching/quick_sort.py b/src/extra_interview_problems/sorting_and_searching/quick_sort.py
new file mode 100644
index 0000000..77034d0
--- /dev/null
+++ b/src/extra_interview_problems/sorting_and_searching/quick_sort.py
@@ -0,0 +1,17 @@
+#!/bin/python
+
+''' Quick sort an array '''
+
+
+def quick_sort(arr):
+    if len(arr) < 2: return arr
+    piv = len(arr)//2
+    left = [x for i, x in enumerate(arr) if x <= arr[piv] and i != piv]
+    right = [x for i, x in enumerate(arr) if x > arr[piv] and i != piv]
+    return quick_sort(left) + [arr[piv]] + quick_sort(right)
+
+
+
+if __name__ == '__main__':
+    arr = [8, 5, 2, 6, 1, 2, 9, 4]
+    print(quick_sort(arr))
\ No newline at end of file
diff --git a/src/useful_scripts/count_unique_words_files.py b/src/useful_scripts/count_unique_words_files.py
new file mode 100755
index 0000000..77ae6b6
--- /dev/null
+++ b/src/useful_scripts/count_unique_words_files.py
@@ -0,0 +1,24 @@
+#!/usr/bin/python
+import collections
+import string
+import sys
+
+def count_unique_word_file():
+    if len(sys.argv) < 2:
+        print "Usage: python count_unique_word.py NAMEFILE"
+
+    words = collections.defaultdict(int)
+    strip = string.whitespace + string.punctuation + string.digits + "\"'"
+    for filename in sys.argv[1:]:
+        with open(filename) as file:
+            for line in file:
+                for word in line.lower().split():
+                    word = word.strip(strip)
+                    if len(word) > 2:
+                        words[word] = +1
+    for word in sorted(words):
+        print("'{0}' occurs {1} times.".format(word, words[word]))
+
+
+if __name__ == '__main__':
+    count_unique_word_file()
\ No newline at end of file