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bt3gl 2023-07-30 21:40:09 -07:00
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#!/usr/bin/env python
__author__ = "bt3"
from collections import Counter
def is_anagram(s1, s2):
'''
>>> is_anagram('cat', 'tac')
True
>>> is_anagram('cat', 'hat')
False
'''
counter = Counter()
for c in s1:
counter[c] += 1
for c in s2:
counter[c] -= 1
for i in counter.values():
if i:
return False
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__':
import doctest
doctest.testmod()

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__author__ = "bt3"
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.

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#!/usr/bin/env python
__author__ = "bt3"
'''
Given a N different open and close braces in a string "( { [ } ] )".
How do you check whether the string has matching braces.
'''
from collections import Counter
def check_if_balance(string):
'''
>>> check_if_balance('{[[(])}]')
True
>>> check_if_balance('{[[()}]')
False
>>> check_if_balance('')
True
'''
table = Counter()
for i in string:
index = str(ord(i))[0]
if i in '{[(':
table[index] += 1
elif i in ')}]':
table[index] -= 1
for i in table.values():
if i !=-0:
return False
return True
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
"""
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.
"""
from collections import defaultdict, Counter
def check_sum(array, k):
'''
>>> check_sum([3, 2, 6, 7, 9, 1], 8)
[(6, 2), (1, 7)]
>>> check_sum([5, 2, 6, 7, 9, 1], 4)
[]
>>>
'''
dict = defaultdict()
res = []
for i in array:
if k-i in dict:
res.append((i, k-i))
del dict[k-i]
else:
dict[i] = 1
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()

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#!/usr/bin/env python
__author__ = "bt3"
''' Determine if an Array of integers contains 3 numbers that sum to 0 '''
from collections import defaultdict
def find_3_number(array):
'''
>>> find_3_number([1,5,56,11,-3,-12])
1 11 -12
True
>>> find_3_number([] )
False
'''
hash_ = defaultdict()
for i in array:
hash_[i] = 1
for i, x in enumerate(array):
for y in array[i+1:]:
if -(x+y) in hash_:
print x, y, -(x+y)
return True
return False
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
given an array of intervals, return max number of non-overlapping intervals
'''
from collections import defaultdict
def non_overlapping(array):
'''
>>> non_overlapping([(1,2), (2,5), (1, 6)])
[[(1, 2), (2, 5)]]
>>> non_overlapping([(1,4), (2,5), (3, 6)])
[]
'''
total = []
for i, t in enumerate(array):
start = t[0]
end = t[1]
for tt in array[i+1:] :
if end <= tt[0]:
total.append([(start, end), (tt[0], tt[1])])
return total
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' convert an integer to a string in any base'''
def convert_from_dec_to_any_base(number, base):
'''
>>> number, base = 9, 2
>>> convert_from_dec_to_any_base(number, base)
'1001'
'''
convertString = '012345679ABCDEF'
if number < base:
return convertString[number]
else:
return convert_from_dec_to_any_base(number//base, base) + \
convertString[number%base]
def convert_from_decimal_to_binary(number, base):
'''
>>> number, base = 9, 2
>>> convert_from_decimal_to_binary(number, base)
1001
'''
multiplier, result = 1, 0
while number > 0:
result += number%base*multiplier
multiplier *= 10
number = number//base
return result
def convert_from_decimal_larger_bases(number, base):
'''
>>> number, base = 31, 16
>>> convert_from_decimal_larger_bases(number, base)
'1F'
'''
strings = "0123456789ABCDEFGHIJ"
result = ""
while number > 0:
digit = number%base
result = strings[digit] + result
number = number//base
return result
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
def conv_int2str(int1):
'''
>>> conv_str2int('367')
367
>>> conv_str2int('0')
0
>>> conv_str2int('-10')
-10
>>> conv_str2int('1e5')
100000
'''
aux_dict = {key:value for key in range(10) for value in '0123456789'[key]}
if int1 == 0:
return '0'
elif int1 < 0:
sign = '-'
int1 = int1*(-1)
else:
sign = ''
aux_ls = []
while int1 > 0:
c = int1%10
int1 = int1//10
cadd = aux_dict[c]
aux_ls.append(cadd)
aux_ls.reverse()
return sign + ''.join(aux_ls)
def conv_str2int(str1):
'''
>>> conv_int2str(0)
'0'
>>> conv_int2str(1e5)
'100000'
>>> conv_int2str(367)
'367'
>>> conv_int2str(-10)
'-10'
'''
if not str1:
return None
aux_dict = {key:value for value in range(10) for key in '0123456789'[value]}
if str1[0] == '-':
sign = -1
str1 = str1[1:]
else:
sign = 1
dec, result = 1, 0
for i in range(len(str1)-1, -1, -1):
aux = str1[i]
if aux == 'e':
exp_num = conv_str2int(str1[i+1:])
number = conv_str2int(str1[:i])
result = number*10**exp_num
break
result += aux_dict[aux]*dec
dec *= 10
return result*sign
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
import collections
import string
import sys
def count_unique_word():
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]))

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#!/usr/bin/env python
__author__ = "bt3"
''' find and delete all the duplicate characters in a string '''
from collections import Counter
def delete_unique(str1):
'''
>>> delete_unique("Trust no one")
'on'
>>> delete_unique("Mulder?")
''
'''
str_strip = ''.join(str1.split())
repeat = Counter()
for c in str_strip:
repeat[c] += 1
result = ''
for c, count in repeat.items():
if count > 1:
result += c
return result
def removing_duplicates_seq(str1):
'''
>>> delete_unique("Trust no one")
'on'
>>> delete_unique("Mulder?")
''
'''
seq = str1.split()
result = set()
for item in seq:
if item not in result:
#yield item
result.add(item)
return result
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__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
def fib_rec(n):
'''
>>> fib_rec(2)
1
>>> fib_rec(5)
5
>>> fib_rec(7)
13
'''
if n < 3:
return 1
return fib_rec(n - 1) + fib_rec(n - 2)
if __name__ == '__main__':
import doctest
doctest.testmod()
fib = fib_generator()
print(next(fib))
print(next(fib))
print(next(fib))
print(next(fib))

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#!/usr/bin/env python
__author__ = "bt3"
def find_0_MxN(m):
''' find 0s in a matrix and replace the col and row to 0s:
>>> m1 = [[1,2,3,4], [5,6,7,8], [9,10,11,12], [13,14,15,16]]
>>> find_0_MxN(m1)
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
>>> m2 = [[1, 2, 3, 4], [0, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
>>> find_0_MxN(m2)
[[0, 2, 3, 4], [0, 0, 0, 0], [0, 10, 11, 12], [0, 14, 15, 16]]
'''
index = []
for row in range(len(m)):
for col in range(len(m[0])):
if m[row][col] == 0:
index.append((row, col))
for i in index:
row = i[0]
col = i[1]
for i in range(len(m)):
m[row][i] = 0
for i in range(len(m[0])):
m[i][col] = 0
return m
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
given 2 dice, determine number of ways to sum S if all dice are rolled
'''
from collections import Counter, defaultdict
def find_dice_probabilities(S=5, n_faces=6):
if S > 2*n_faces or S < 2:
return None
cdict = Counter()
ddict = defaultdict(list)
for dice1 in range(1, n_faces+1):
for dice2 in range(1, n_faces+1):
t = [dice1, dice2]
cdict[dice1+dice2] += 1
ddict[dice1+dice2].append( t)
return [cdict[S], ddict[S]]
if __name__ == '__main__':
results = find_dice_probabilities()
assert(results[0] == len(results[1]))

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#!/usr/bin/env python
__author__ = "bt3"
''' computes the edit distance between two strings '''
def find_edit_distance(str1, str2):
'''
>>> s = 'sunday'
>>> t = 'saturday'
>>> find_edit_distance(s, t)
3
'''
m = len(str1)
n = len(str2)
diff = lambda c1, c2: 0 if c1 == c2 else 1
E = [[0] * (n + 1) for i in range(m + 1)]
for i in range(m + 1):
E[i][0] = i
for j in range(1, n + 1):
E[0][j] = j
for i in range(1, m + 1):
for j in range(1, n + 1):
E[i][j] = min(E[i-1][j] + 1, E[i][j-1] + 1, E[i-1][j-1] + diff(str1[i-1], str2[j-1]))
return E[m][n]
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
from collections import Counter
def find_non_rep_char(s1):
'''
>>> s1 = 'aabbcceff'
>>> find_non_rep_char(s1)
e
>>> find_non_rep_char('ccc')
'''
aux_dict = Counter()
for i in s1:
aux_dict[i] += 1
for k, v in aux_dict.items():
if v < 2:
print k
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
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()

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#!/usr/bin/env python
__author__ = "bt3"
def find_substr(s1, s2):
if len(s1) < len(s2):
bs = s2
ss = s1
else:
bs = s1
ss = s2
ps = 0
for c in bs:
if ss[ps] == c:
ps += 1
else:
ps = 0
if ps == len(ss)-1:
return True
return False
if __name__ == '__main__':
s1 = 'buffy is a vampire slayer'
s2 = 'vampire'
s3 = 'angel'
assert(find_substr(s2, s1) == True)
assert(find_substr(s3, s1) == False)

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#!/usr/bin/env python
__author__ = "bt3"
import collections
def find_if_unique_chars(word):
"""
>>> find_if_unique_chars('abcde')
True
>>> find_if_unique_chars('abcae')
False
"""
unique = True
counter = collections.Counter()
for c in word:
if not counter[c]:
counter[c] += 1
else:
unique = False
return unique
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
You are given an array of integers (both positive and negative).
Find the contiguous sequence with the largest sum.
'''
def find_largest_sum(array):
'''
>>> find_largest_sum([-1, 2, -3, 5, 3, 1, -16, 7, 1, -13, 1])
9
>>> find_largest_sum([])
0
>>> find_largest_sum([1])
1
'''
sum_ = 0
sum_here = 0
for i in array:
sum_here += i
if sum_here < 0:
sum_here = 0
if sum_ < sum_here:
sum_ = sum_here
return sum_
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' find the longest continuous increasing subsequence'''
def find_long_con_inc(seq):
'''
>>> find_long_con_inc([1, -2, 3, 5, 1, -1, 4, -1, 6])
[-2, 3, 5]
>>> find_long_con_inc([1, 3, -2, 3, 5, 6])
[-2, 3, 5, 6]
>>> find_long_con_inc([1, 3, 4, -13, 2, 5, 8, -1, 2,-17])
[-13, 2, 5, 8]
'''
res, aux = [], []
seq.append(-float('infinity'))
for i, n in enumerate(seq[:-1]):
aux.append(n)
if n > seq[i+1]:
if len(res) < len(aux):
res = aux[:]
aux = []
return res
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
given a string, find longest string with unique characters
'''
def find_longest(string):
'''
>>> find_longest('abfgrhgtrdsandwejfhdasjcbdsjvrejwghireeej')
'wejfhdas'
>>> find_longest('abcabcabcabcdefabcccc')
'defabc'
'''
maxs = ''
result = ''
for c in string:
if c in result:
if len(maxs) < len(result):
maxs = result
result = ''
else:
result += c
if result and len(maxs) < len(result):
maxs = result
return maxs
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
from collections import defaultdict
def find_unique_number(array):
'''
>>> find_unique_number([1, 3, 6, 1, 5, 6, 9, 3, 7])
[1, 6, 3]
>>> find_unique_number([1, 3, 5, 6, 9, 7])
[]
'''
table = defaultdict()
total = []
for i in array:
if i in table:
total.append(i)
else:
table[i] = 1
return total
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''Given an array of numbers, replace each number with the product of all
the numbers in the array except the number itself *without* using division
'''
def find_product_without_division(seq):
'''
>>> seq = [2,3,4]
>>> find_product_without_division(seq)
[12, 8, 6]
'''
forw = []
bacw = []
for i in range(len(seq)):
prod_f, prod_b = 1, 1
for next in range(i+1, len(seq)):
prod_f *= seq[next]
for before in range(0, i):
prod_b *= seq[before]
forw.append(prod_f)
bacw.append(prod_b)
for i in range(len(seq)):
seq[i] = forw[i] * bacw[i]
return seq
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
from collections import Counter
def find_top_N_recurring_words(seq, N):
''' find the top N recurring words in a file:
1) use a hash table to find counts
2) sort the list on base of the maximum counts
3) return the last N words
'''
dcounter = Counter()
for word in seq.split():
dcounter[word] += 1
return dcounter.most_common(N)
if __name__ == '__main__':
seq = 'buffy angel monster xander a willow gg buffy the monster super buffy angel'
N = 3
assert(find_top_N_recurring_words(seq, N) == [('buffy', 3), ('monster', 2), ('angel', 2)])

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#!/usr/bin/env python
__author__ = "bt3"
"""
Two numbers out of n numbers from 1 to n are missing.
The remaining n-2 numbers are in the array but not sorted.
Find the missing numbers The sum1 is the sum of all the elements in n.
The sum2 is the sum of all the elements in n-2. sum1 - sum2 = num1 + num2 = s.
The prod1 is the prod of all the elements in n. The prod2 is the prod of all
the elements in n-2. prod1/prod2 = num1*num2 =p.
Runtime is O(n), because it scan 3 times. Space is O(1)
In case of finding one integer, Let the missing number be M. We know that
the sum of first N natural numbers is N*(N+1)/2. Traverse through the array
once and calculate the sum. This is the sum of first N natural numbers -
M or S=N*(N+1)/2 - M. Therefore M = N*(N+1)/2 - S.
"""
import math
def find_two_missing_numbers(l1):
'''
>>> l1 = [1, 3, 5]
>>> find_two_missing_numbers(l1)
(4, 2)
'''
n_min_2 = len(l1)
n = n_min_2 + 2
sum1, sum2, prod1, prod2 = 0, 0, 1, 1
sum2 = sum(l1[:])
sum1 = sum(range(1,n+1))
s = sum1 - sum2
for i in range(1, n-1):
prod1 = prod1*i
prod2 = prod2*l1[i-1]
prod1 = prod1*n*(n-1)
p = prod1/prod2
num1 = (s + math.sqrt(s*s - 4*p))/2
num2 = (s - math.sqrt(s*s - 4*p))/2
return int(num1), int(num2)
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' 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 exit 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'

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#!/usr/bin/env python
__author__ = "bt3"
def intersection_two_arrays_sets(seq1, seq2):
'''
>>> intersection_two_arrays_sets([1,2,3,5,7,8], [3,5,6])
set([3, 5])
>>> intersection_two_arrays_sets([1,2,7,8], [3,5,6])
set([])
'''
# O(n)
set1 = set(seq1)
set2 = set(seq2)
return set1.intersection(set2) #same as list(set1 & set2)
def intersection_two_arrays_On2(seq1, seq2):
'''
>>> intersection_two_arrays_On2([1,2,3,5,7,8], [3,5,6])
[3, 5]
>>> intersection_two_arrays_On2([1,2,7,8], [3,5,6])
[]
'''
final = []
for i in seq1:
for j in seq2:
if i == j:
final.append(i)
return final
def intersection_two_arrays_On(seq1, seq2):
'''
>>> intersection_two_arrays_On([1,2,3,5,7,8], [3,5,6])
[5, 3]
>>> intersection_two_arrays_On([1,2,7,8], [3,5,6])
[]
'''
final = []
while seq1 and seq2:
if seq1[-1] == seq2[-1]:
final.append(seq1.pop())
seq2.pop()
elif seq1[-1] > seq2[-1]:
seq1.pop()
else:
seq2.pop()
return final
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
def beating_stock(array):
imin = 0
# first deal is just buying in the next day (1)
deal = [array[1] - array[imin], imin, 1]
for i, d in enumerate(array):
deal_here = d - array[imin]
if deal_here > deal[0]:
deal = [deal_here, imin, i]
elif d < array[imin]:
imin = i
return deal[0], array[deal[1]], array[deal[2]]
def beating_stock2(array):
deal = 0
min_value = array[0]
for i, d in enumerate(array):
deal_here = d - min_value
if deal_here > deal:
deal = deal_here
else:
if min_value > array[i]:
min_value = array[i]
return deal
if __name__ == '__main__':
array = [7, 2, 3, 6, 5, 8, 5, 3, 4]
deal = beating_stock(array)
print("Profit: %d, buying at %d, selling at %d." %(deal[0], deal[1], deal[2]))
deal = beating_stock2(array)
print "Profit: " + str(deal)

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#!/usr/bin/env python
__author__ = "bt3"
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

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#!/usr/bin/env python
__author__ = "bt3"
from collections import defaultdict
def is_palindrome(array):
'''
>>> is_palindrome('subi no onibus')
True
>>> is_palindrome('helllo there')
False
>>> is_palindrome('h')
True
>>> is_palindrome('')
True
'''
array = array.strip(' ')
if len(array) < 2:
return True
if array[0] == array[-1]:
return is_palindrome(array[1:-1])
else:
return False
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
def perm(str1):
'''
>>> perm('123')
['123', '132', '231', '213', '312', '321']
'''
if len(str1) < 2:
return str1
res = []
for i, c in enumerate(str1):
for cc in perm(str1[i+1:] + str1[:i]):
res.append(c + cc)
return res
def perm2(str1):
'''
>>> perm2('123')
['123', '132', '213', '231', '312', '321']
'''
from itertools import permutations
return [''.join(p) for p in permutations(str1)]
def ispermutation(s1, s2):
'''
>>> ispermutation('231', '123')
True
>>> ispermutation('231', '153')
False
'''
from collections import Counter
aux = Counter()
for i in s1:
aux[i] += 1
for i in s2:
aux[i] -= 1
for v in aux.values():
if v != 0:
return False
return True
def ispermutation2(s1, s2):
'''
>>> ispermutation2('231', '123')
True
>>> ispermutation2('231', '153')
False
'''
if sorted(s1) == sorted(s2):
return True
else:
return False
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
''' Write code to generate all possible case permutations of a given
lower-cased string
'''
def alpha_permutation(string):
'''
>>> alpha_permutation('0ab')
['0Ab', '0Ab', '0ab', '0ab', '0Ba', '0Ba', '0ba', '0ba', 'ab0', 'a0b', 'a0b', 'b0a', 'b0a', 'ba0']
>>> alpha_permutation('')
''
'''
if len(string) < 2:
return string
result = []
for i, c in enumerate(string):
rest = string[i+1:] + string[:i]
for cc in alpha_permutation(rest):
if cc.isalpha():
result.append(c.upper() + cc)
result.append(c + cc)
return result
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
find prime factors of a number.
'''
import math
import random
def find_prime_factors(n):
'''
>>> find_prime_factors(14)
[2, 7]
>>> find_prime_factors(19)
[]
'''
divisors = [d for d in range(2, n//2 + 1) if n % d == 0]
primes = [d for d in divisors if is_prime(d)]
return primes
def is_prime(n):
for j in range(2, int(math.sqrt(n))):
if (n % j) == 0:
return False
return True
''' return a n-bit prime '''
def generate_prime(number=3):
while 1:
p = random.randint(pow(2, number-2), pow(2, number-1)-1)
p = 2 * p + 1
if find_prime_factors(p):
return p
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
def get_products_of_all_except_at_index(array):
'''
>>> a = [1, 7, 3, 4]
>>> get_products_of_all_except_at_index(a)
[84, 12, 28, 21]
'''
total = 1
for n in array:
total *= n
new_array = []
for n in array:
if n is not 0:
item = total/n
new_array.append(item)
else:
new_array.append(n)
return new_array
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
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)
if __name__ == '__main__':
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))

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#!/usr/bin/env python
__author__ = "bt3"
def revert(string):
'''
>>> s = 'hello'
>>> revert(s)
'olleh'
>>> revert('')
''
'''
return string[::-1]
def reverse_string_inplace(s):
'''
>>> s = 'hello'
>>> reverse_string_inplace(s)
'olleh'
>>> reverse_string_inplace('')
''
'''
if s:
s = s[-1] + reverse_string_inplace(s[:-1])
return s
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
def reversing_words(word):
"""
>>> reversing_words('buffy is awesome')
'awesome is buffy'
"""
new_word = []
words = word.split(' ')
for word in words[::-1]:
new_word.append(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()

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#!/usr/bin/env python
__author__ = "bt3"
def rotate_NxN(m):
n = len(m)
for layer in range(n//2):
first = layer
last = n - 1 - layer
for i in range(first, last):
offset = i - first
top = m[first][i]
m[first][i] = m[last-offset][first]
m[last-offset][first] = m[last][last-offset]
m[last][last-offset] = m[i][last]
m[i][last] = top
return m
def main():
m = [[1,2],[3,4]]
mr = [[3,1],[4,2]]
assert(rotate_NxN(m) == mr)
m2 = [[1,2,3],[4,5,6],[7,8,9]]
print(rotate_NxN(m2))
if __name__ == '__main__':
main()

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#!/usr/bin/env python
__author__ = "bt3"
'''To use timeit you create a Timer object whose parameters are two Python
statements. The first parameter is a Python statement that you want to time;
the second parameter is a statement that will run once to set up the test.
The timeit module will then time how long it takes to execute the statement
some number of times. By default timeit will try to run the statement one
million times. When its done it returns the time as a floating point value
representing the total number of seconds. However, since it executes the
statement a million times you can read the result as the number of
microseconds to execute the test one time. You can also pass timeit a
named parameter called number that allows you to specify how many times the
test statement is executed. The following session shows how long it takes to
run each of our test functions 1000 times. '''
import timeit
import random
for i in range(10000,1000001,20000):
''' this experiment confirm that the contains operator for lists is O(n) and for dict is O(1) '''
t = timeit.Timer("random.randrange(%d) in x"%i, "from __main__ import random,x")
x = list(range(i))
lst_time = t.timeit(number=1000)
x = {j:None for j in range(i)}
d_time = t.timeit(number=1000)
print("%d,%10.3f,%10.3f" % (i, lst_time, d_time))
""" There results are:
10000, 0.192, 0.002
30000, 0.600, 0.002
50000, 1.000, 0.002
70000, 1.348, 0.002
90000, 1.755, 0.002
110000, 2.194, 0.002
130000, 2.635, 0.002
150000, 2.951, 0.002
170000, 3.405, 0.002
190000, 3.743, 0.002
210000, 4.142, 0.002
230000, 4.577, 0.002
250000, 4.797, 0.002
270000, 5.371, 0.002
290000, 5.690, 0.002
310000, 5.977, 0.002
so we can see the linear tile for lists, and constant for dict!
Big-O Efficiency of Python Dictionary Operations
Operation Big-O Efficiency
copy O(n)
get item O(1)
set item O(1)
delete item O(1)
contains (in) O(1)
iteration O(n)
"""

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#!/usr/bin/env python
__author__ = "bt3"
'''To use timeit you create a Timer object whose parameters are two Python statements. The first parameter is a Python statement that you want to time; the second parameter is a statement that will run once to set up the test. The timeit module will then time how long it takes to execute the statement some number of times. By default timeit will try to run the statement one million times. When its done it returns the time as a floating point value representing the total number of seconds. However, since it executes the statement a million times you can read the result as the number of microseconds to execute the test one time. You can also pass timeit a named parameter called number that allows you to specify how many times the test statement is executed. The following session shows how long it takes to run each of our test functions 1000 times. '''
def test1():
l = []
for i in range(1000):
l = l + [i]
def test2():
l = []
for i in range(1000):
l.append(i)
def test3():
l = [i for i in range(1000)]
def test4():
l = list(range(1000))
if __name__ == '__main__':
import timeit
t1 = timeit.Timer("test1()", "from __main__ import test1")
print("concat ",t1.timeit(number=1000), "milliseconds")
t2 = timeit.Timer("test2()", "from __main__ import test2")
print("append ",t2.timeit(number=1000), "milliseconds")
t3 = timeit.Timer("test3()", "from __main__ import test3")
print("comprehension ",t3.timeit(number=1000), "milliseconds")
t4 = timeit.Timer("test4()", "from __main__ import test4")
print("list range ",t4.timeit(number=1000), "milliseconds")
""" The results are:
('concat ', 2.366791009902954, 'milliseconds')
('append ', 0.16743111610412598, 'milliseconds')
('comprehension ', 0.06446194648742676, 'milliseconds')
('list range ', 0.021029949188232422, 'milliseconds')
So we see the following pattern for lists:
Operation Big-O Efficiency
index [] O(1)
index assignment O(1)
append O(1)
pop() O(1)
pop(i) O(n)
insert(i,item) O(n)
del operator O(n)
iteration O(n)
contains (in) O(n)
get slice [x:y] O(k)
del slice O(n)
set slice O(n+k)
reverse O(n)
concatenate O(k)
sort O(n log n)
multiply O(nk)
"""

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#!/usr/bin/env python
__author__ = "bt3"
''' Search an Entry in a Matrix where the Rows and columns are Sorted
In this 2D matrix, every row is increasingly sorted from left to right,
and every column is increasingly sorted from top to bottom.
The runtime is O(m+n).
'''
def find_elem_matrix_bool(m1, value):
found = False
row = 0
col = len(m1[0]) - 1
while row < len(m1) and col >= 0:
if m1[row][col] == value:
found = True
break
elif m1[row][col] > value:
col-=1
else:
row+=1
return found
def test_find_elem_matrix_bool(module_name='this module'):
m1 = [[1,2,8,9], [2,4,9,12], [4,7,10,13], [6,8,11,15]]
assert(find_elem_matrix_bool(m1,8) == True)
assert(find_elem_matrix_bool(m1,3) == False)
m2 = [[0]]
assert(find_elem_matrix_bool(m2,0) == True)
if __name__ == '__main__':
test_find_elem_matrix_bool()

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#!/usr/bin/env python
__author__ = "bt3"
from collections import Counter
def str_comp(s):
'''
>>> s1 = 'aabcccccaaa'
>>> str_comp(s1)
'a2b1c5a3'
>>> str_comp('')
''
'''
count, last = 1, ''
list_aux = []
for i, c in enumerate(s):
if last == c:
count += 1
else:
if i != 0:
list_aux.append(str(count))
list_aux.append(c)
count = 1
last = c
list_aux.append(str(count))
return ''.join(list_aux)
if __name__ == '__main__':
import doctest
doctest.testmod()

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#!/usr/bin/env python
__author__ = "bt3"
'''
find the sum of two integers represented as strings,
return the sum as string, i.e "123" and "456" would return "579"
'''
def get_number(s):
count = 1
num = 0
p = len(s) -1
while p >= 0:
num = num + int(s[p])*count
count *= 10
p -= 1
return num
def sum_string(s1, s2):
'''
>>> sum_string('10', '5')
'15'
>>> sum_string('0', '1')
'1'
>>> sum_string('123', '456')
'579'
'''
n1 = get_number(s1)
n2 = get_number(s2)
return str(n2 + n1)
if __name__ == '__main__':
import doctest
doctest.testmod()