Coding Questions V - 2023

What is Hashable?

Let's start with the key in dictionary. The key of a dictionary should be unique. Internally, the key is transformed by a hash function and become an index to a memory where the value is stored. Suppose, if the key is changed, then it will be pointing somewhere else and we lose the original value the the old key was pointing.

In Python, we can say that the key should be hashable. That's why we always use hashable objects as keys. Hashable objects are integers, floats, strings, tuples, and frozensets.

Immutable object will not change after it has been created, especially in any way that changes the hash value of that object. Objects used as hash keys must be immutable so their hash values remain unchanged. Note that all hashable objects are also immutable objects.

Mutable objects are lists, dictoroaries, and sets.

Don't be surprised at the following code using tuple as a key for a dictionary:

d={}

t1 = (1,2)
t2 = (3,4,5)

d[t1] = 2
d[t2] = 3

print d[t1], d[t2]

Output:

2 3

We rarely use a tuple as a key, but there is no problem because tuple is hashable.

If we use a list as a key, we get the following error:

print d[list(range(2))]

TypeError: unhashable type: 'list'

An object is hashable if it has a hash value which never changes during its lifetime (it needs a __hash__() method), and can be compared to other objects (it needs an __eq__() or __cmp__() method). Hashable objects which compare equal must have the same hash value.

Hashability makes an object usable as a dictionary key and a set member, because these data structures use the hash value internally.

All of Python's immutable built-in objects are hashable, while no mutable containers (such as lists or dictionaries) are. Objects which are instances of user-defined classes are hashable by default; they all compare unequal, and their hash value is their id().

-from Python glossary

t = tuple([0,1,2,3])
print(t)

Output:

(0, 1, 2, 3)

# list of strings
L = ["Imagination", "is", "more", "important", "than", "knowledge"]
print(' '.join(L))

# list of integers
N = [0,1,2,3]
print(''.join(str(n) for n in N))

Output:

Imagination is more important than knowledge
0123

number = ['eins','zwei','drei',1,2,3]
print(dict(zip(number[:3], number[3:])))  

Output:

{'drei': 3, 'eins': 1, 'zwei': 2}

number = ['eins',1,'zwei',2,'drei',3]
print(dict(zip(number[0::2], number[1::2]))) 

Specially, for this zigzag (on-and-off) type, we can use iter() method, and create list iterator object:

number = ['eins',1,'zwei',2,'drei',3]
i = iter(number)
print(type(i))
print(dict(zip(i,i)))  

We'll get the same output:

<type 'listiterator'>
{'drei': 3, 'eins': 1, 'zwei': 2}

While the append() and extend() methods modify the original list, the concatenate ('+') creates a new list:

a = [1,2]
b = [3,4,5]
c = a.append(b) # modifying a
print(a, c)

a = [1,2]
b = [3,4,5]
c = a.extend(b) # modifying a
print(a, c)

a = [1,2]
b = [3,4,5]
c = a + b  # creating a new list
print(a, c)

Output:

[1, 2, [3, 4, 5]] None
[1, 2, 3, 4, 5] None
[1, 2] [1, 2, 3, 4, 5]

The sorted() function returns a new sorted list.

We can also use the list.sort() method. It modifies the list in-place (and returns None to avoid confusion). Usually it's less convenient than sorted() - but if we don't need the original list, list.sort() is slightly more efficient.

a = [3, 2, 1]

a2 = sorted(a)
print(a, a2)  

a3 = a.sort()
print(a, a3)

Output:

[3, 2, 1] [1, 2, 3]
[1, 2, 3] None

The list.sort() method is only defined for lists. In contrast, the sorted() function accepts any iterable.

Note that both list.sort() and sorted() have a key parameter to specify a function (or other callable) to be called on each list element prior to making comparisons:

from operator import itemgetter, attrgetter

guest = [
    ('john', 'A', 15, 170),
    ('clair', 'B', 44, 240),
    ('bell', 'D', 74, 111),
    ('dave', 'C', 10, 164) ]

print(sorted(guest, key = lambda x:x[0]))
print(sorted(guest, key = itemgetter(1)))

class Guest:
  def __init__(self, name, grade, age, weight):
    self.name = name
    self.grade = grade
    self.age = age
    self.weight = weight
  def __repr__(self):
    return repr( (self.name, self.grade, self.age, self.weight) )

g_object = [
    Guest('john', 'A', 15, 170),
    Guest('clair', 'B', 44, 240),
    Guest('bell', 'D', 74, 111),
    Guest('dave', 'C', 10, 164),
 ]

print(sorted(g_object, key = lambda g: g.age))
print(sorted(g_object, key = attrgetter('weight')))

Output:

[('bell', 'D', 74, 111), ('clair', 'B', 44, 240), ('dave', 'C', 10, 164), ('john', 'A', 15, 170)]
[('john', 'A', 15, 170), ('clair', 'B', 44, 240), ('dave', 'C', 10, 164), ('bell', 'D', 74, 111)]
[('dave', 'C', 10, 164), ('john', 'A', 15, 170), ('clair', 'B', 44, 240), ('bell', 'D', 74, 111)]
[('bell', 'D', 74, 111), ('dave', 'C', 10, 164), ('john', 'A', 15, 170), ('clair', 'B', 44, 240)]   

We have a list : [1,1,2,3,4,5,6,6,6]. Print how many 6s in the list in one-liner:

print([1,1,2,3,4,5,6,6,6].count(6))

There is a Counter() method which is faster:

from collections import Counter
L = [1,1,2,3,4,5,6,6,6]
print(Counter(L))

Output:

Counter({6: 3, 1: 2, 2: 1, 3: 1, 4: 1, 5: 1})

So, we can pass the number as a key to the dictionary to get its count. For example, pass 6 as a key:

print Counter([1,1,2,3,4,5,6,6,6])[6]  # we get 3

We can write a simple version of the Counter() like this:

def myCounter(L):
  import operator
  # initialize dict with 0
  d = dict.fromkeys(L,0)
  # or 
  # d = {x:0 for x in L}
  for l in L:
    d[l] += 1
  # return sorted (by vaule) dict descending order
  return sorted(d.items(), key=operator.itemgetter(1), reverse=True)
  # or using labmda
  # return sorted(d.items(), key = lambda x:x[1], reverse=True)

L = [1,1,2,3,4,5,6,6,6]
print(myCounter(L))    

Given two lists, a = ['l','m','n'] and b = ['p','q','r']. Write a code to crete a list, ['lp', 'mq', 'nr'].

As a preview, the following code shows how the zip(list,list) works:

#!/usr/local/bin/python

a = ['l','m','n']
b = ['p','q','r']
c = [(x,y) for x,y in zip(a,b)]
print(c)    # ==> [('l', 'p'), ('m', 'q'), ('n', 'r')]
d = [[x,y] for x,y in zip(a,b)]
print(d)    # ==> [['l', 'p'], ['m', 'q'], ['n', 'r']]

Our code should look like this:

#!/usr/local/bin/python

j = [''.join([x,y]) for x,y in zip(a,b)]
# or
# j = [x+y for x,y in zip(a,b)]
print(j)

Output:

['lp', 'mq', 'nr']    

Note that join() takes a list!

We have two lists: [values] and [weights]. This sample shows how to use zip to calculate weighted average:

weights = [0.5, 0.8, 0.7, 0.8, 1.0]
values = [10, 20, 10, 10, 30]
average = sum([weight[i] * values[i] for i in range(len(weight))])/sum(values)
zip_average = sum ([x * y for x,y in zip(weights,values)]) / sum(values)
print average, zip_average  # 0.825 0.825

Note that the sum(list) takes a list, and the zip(list,list) takes two lists.

Here we can use two approach: filter() and list comprehension:

a = [1,2,3,5,7,9]
b = [2,3,5,6,7,8]
print filter(lambda x: x in a, b)  # prints out [2, 3, 5, 7]

Note that we can do the same with list comprehension:

a = [1,2,3,5,7,9]
b = [2,3,5,6,7,8]
print [x for x in a if x in b] # prints out [2, 3, 5, 7]

With set():

a = [1,2,3,5,7,9]
b = [2,3,5,6,7,8]

j = list(set(a) & set(b))
print(j)  # ==> [2, 3, 5, 7]  

In this section, initially a list of character is given. From the list, we make a dictionary {char, count}, and print out the dictionary sorted by the value (count).

a = ['a','a','c','c','b','c']
# d = dict.fromkeys(a,0)
# or
d = {x:0 for x in a}
print(d)

for ch in a:
	d[ch] += 1
print(d)

sorted_d = sorted(d.items(), key=lambda x:x[1], reverse=True)
print(sorted_d)

Output:

{'a': 0, 'c': 0, 'b': 0}
{'a': 2, 'c': 3, 'b': 1}
[('c', 3), ('a', 2), ('b', 1)]

We can also use operator.itemgetter() to sort dictionary:

inventory = {'apple': 3, 'banana': 2, 'pear': 5, 'orange': 1}
s = sorted(inventory.items(), key=operator.itemgetter(0))
print(s)

Output:

[('apple', 3), ('banana', 2), ('orange', 1), ('pear', 5)]

The file given: "input.txt" is "abcdefgHIJKLMOPQrstuvwxyz". We need to count the Capitals only. One line of code:

print(len([s for s in open("input.txt","r").read() if s.isupper()]))

Output:

9

Actually, more readable code looks something like this:

with open("input.txt","r") as f:
    print(len([s for s in f.read() if s.isupper()]))

Write a program that computes all Armstrong numbers in the range of 0 and 999. An Armstrong number is a number such that the sum of its digits raised to the third power is equal to the number itself. For example, 371 is an Armstrong number, since 3**3 + 7**3 + 1**3 = 371.

Hint: use floor (//)

def isArmstrong(n):
    temp = n
    sum = 0
    while temp > 0:
        sum += (temp % 10)**3
        temp //= 10
    if n == sum:
        return True
    return False

n = 1000
print([i for i in range(0,n) if isArmstrong(i)])

Output:

[0, 1, 153, 370, 371, 407]

def GCF(n1,n2):
    n = min(n1,n2)
    while True:
        if n1 % n == 0 and n2 % n == 0:
            return n
        n -= 1

n1 = 80; n2 = 240
print(GCF(n1,n2))

Output:

80

def LCM(n1,n2):
    n = max(n1,n2)
    while True:
        if n % n1 == 0 and n % n2 == 0:
            return n
        n += 1

n1 = 24; n2 = 36
print(LCM(n1,n2))

Output:

72

Write a code to shuffle a deck of card and draw 5 cards.

import itertools, random

deck = list(itertools.product(range(1,14),['Space','Diamond','Club','Heart']))
random.shuffle(deck)
hands = list(deck[i] for i in range(5))
print(hands)

Output:

[(2, 'Club'), (11, 'Diamond'), (8, 'Diamond'), (13, 'Heart'), (2, 'Diamond')]

Example:
# 'AAAABBBCCDAABBB' --> [A, B, C, D] (Ever seen)
# 'AAAABBBCCDAABBB' --> [A, B, C, D, A, B] (Just seen)

s1 = 'AAAABBBCCDAABBB'
s2 = 'AAAABBBCCDAABBB'

ever_seen = [s1[0]]
for v in s1[1:]:
    if v != ever_seen[-1] and v not in ever_seen:
        ever_seen.append(v)
print(ever_seen)


just_seen = [s2[0]]
for v in s2[1:]:
    if v != just_seen[-1]:
        just_seen.append(v)
print(just_seen)

Output:

['A', 'B', 'C', 'D']
['A', 'B', 'C', 'D', 'A', 'B']

Given a list, list items as the most frequent at top:

mylist = ['a','b','b','d','b','c','d','a','d','d']
    
d = {}.fromkeys(mylist,0) 
# or
d = {k:0 for k in mylist}
for m in mylist:
    d[m] += 1
print('initial d = %s' %d)

import operator

# descending : reverse=True
d2 = sorted(d.items(), key=operator.itemgetter(1), reverse=True)
print('order by value: %s' %d2)

d2 = sorted(d.items(), key=operator.itemgetter(0), reverse=True)
print('order by key: %s' %d2)

Output:

initial d = {'a': 2, 'b': 3, 'd': 4, 'c': 1}
order by value: [('d', 4), ('b', 3), ('a', 2), ('c', 1)]
order by key: [('d', 4), ('c', 1), ('b', 3), ('a', 2)]

Note: the itemgetter() can be used like this (specifying the sort order):

import operator

st = [('Tom',19,80), ('John',20,90), ('Jony',17,91), ('Jony',17,93), ('Json',21,85)]
print(sorted(st, key = operator.itemgetter(0,1,2)))

Output:

[('John', 20, 90), ('Jony', 17, 91), ('Jony', 17, 93), ('Json', 21, 85), ('Tom', 19, 80)]

Take an input string parameter and determine if exactly 3 question marks exist between every pair of numbers that add up to 10. If so, return true, otherwise return false.

'''
"arrb6???4xxbl5???eee5" => true
"4???1??bc9" => false
"acc?7??sss?3rr4??????5" => true
"5??aaaaaaaaaaaaaaaaaaa?5?5" => false
"9b???1???9??k?1???9" => true
"3?6???4?j?1???9" => true
"aa6?9" => false
'''

sslist = ["arrb6???4xxbl5???eee5", "4???1??bc9", "acc?7??sss?3rr4??????5",
         "5??aaaaaaaaaaaaaaaaaaa?5?5", "9b???1???9??k?1???9", "3?6???4?j?1???9", "aa6?9"]

def qmark(ss):
    print(ss)
    
    # index : the location of digits
    index = [i for i, s in enumerate(ss) if s.isdigit()]
    print('index=%s' %index)
    
    # index_tuple : [(start1, end1), ... ]
    index_tuple = [(index[i], index[i+1]) for i in range(len(index)-1)]
    print('index_tuple=%s' %index_tuple)

    # truth : list of 'True' or 'False'. 
    # To pass, all items should be 'True', Empty list means failure
    truth = []
    for t in index_tuple:
        sum = int(ss[t[0]]) + int(ss[t[1]])
        if sum == 10:
            if ss[t[0]:t[1]].count("?") != 3:
                truth.append('F')
            else:
                truth.append('T')
    if len(truth) == 0:
        return False
    elif 'F' in truth:
        return False
    else:
        return True
                    
for ss in sslist:
    print(qmark(ss))
    print()

The output:

arrb6???4xxbl5???eee5
index=[4, 8, 13, 20]
index_tuple=[(4, 8), (8, 13), (13, 20)]
True

4???1??bc9
index=[0, 4, 9]
index_tuple=[(0, 4), (4, 9)]
False

acc?7??sss?3rr4??????5
index=[4, 11, 14, 21]
index_tuple=[(4, 11), (11, 14), (14, 21)]
True

5??aaaaaaaaaaaaaaaaaaa?5?5
index=[0, 23, 25]
index_tuple=[(0, 23), (23, 25)]
False

9b???1???9??k?1???9
index=[0, 5, 9, 14, 18]
index_tuple=[(0, 5), (5, 9), (9, 14), (14, 18)]
True

3?6???4?j?1???9
index=[0, 2, 6, 10, 14]
index_tuple=[(0, 2), (2, 6), (6, 10), (10, 14)]
True

aa6?9
index=[2, 4]
index_tuple=[(2, 4)]
False

We can split a string with string.split(). But when we need to split a string with multiple delimiters, we may want to use regular expression, re.split().

For example with the following string:

"Is life a joke?
Does it not make you cry?
Life is not a joke!
Those who live know.
It is sometimes a yoke!"

We want to split the string with three delimiters: '?', '!', and '.'

The string now has multiple sentences.

We should write a code the maximum number of words for those sentences.

Here is the code:

def solution(S):
    import re

    sentences = re.split(r'[.!?]', S)
    print(sentences)

    mx = 0 
    for sn in sentences:
        mx = max (mx, len(sn.split()))
    return mx
       
S = '''Is life a joke? Does it not make you cry? Life is not a joke! My be or may be not at all.'''
print(solution(S))

Output:

['Is life a joke', ' Does it not make you cry', ' Life is not a joke', ' My be or may be not at all', '']
8

Note that in the re.split() we used r (raw) which signals Python not to interpret backslashes and special meta-characters in the string (literally, ignore if there are any special character, for example, dot(.), meta character). Also, we used square bracket, '[]', to tell Python to match any characters in the bracket.

Another example for a given string: '1. First part 2. Second part 3. Third part'. How to split the string to ['First part', 'Second part', 'Third part']?

input_string = '1. First part 2. Second part 3. Third part'
strings = re.split('\d\. ', input_string)
print(strings)

Output:

['', 'First part ', 'Second part ', 'Third part']

Both method accept pattern and string and returns a match object on success or None if no match is found.

The following code prints out the 'count' for a given pattern, 'back':

import re
filename = ['backup.txt','text.back','backup2.txt', 'file.back']
pattern = r'back'
count = 0
for file in filename:
    if re.match(pattern, file):
        print('matched: %s' %file)
        count += 1
print('count=%d' %count)

Output:

matched: backup.txt
matched: backup2.txt
count=2

Note that re.match() is very similar to re.search(), and the difference is that re.match() will start looking for matches at the beginning of the string.

So, we can get the same result from re.search() if we use ^ before the pattern:

import re
filename = ['backup.txt','text.back','backup2.txt', 'file.back']
pattern = r'back'
count = 0
print('pattern=%s' %pattern)
for file in filename:
    if re.match(pattern, file):
        print('re.match() - matched: %s' %file)
        count += 1
print('count=%d' %count)

pattern = r'back'
count = 0
print('pattern=%s' %pattern)
for file in filename:
    if re.search(pattern, file):
        print('re.search() - matched: %s' %file)
        count += 1
print('count=%d' %count)

pattern = r'^back'
count = 0
print('pattern=%s' %pattern)
for file in filename:
    result = re.search(pattern, file)
    if result:
        print('re.search() - matched: %s' %result.group())
        count += 1
print('count=%d' %count)

Output:

pattern=back
re.match() - matched: backup.txt
re.match() - matched: backup2.txt
count=2
pattern=back
re.search() - matched: backup.txt
re.search() - matched: text.back
re.search() - matched: backup2.txt
re.search() - matched: file.back
count=4
pattern=^back
re.search() - matched: back
re.search() - matched: back
count=2

The match object has group() method which contains the matching text in the string.

'''
1. At least 1 letter between [a-z]
2. At least 1 number between [0-9]
1. At least 1 letter between [A-Z]
3. At least 1 character from [@#$%^&+=]
4. Minimum length of transaction password: 6
'''

import re

def pass_check(password):
    if re.match(r"^(?=.*[\d])(?=.*[A-Z])(?=.*[a-z])(?=.*[@#$])[\w\d@#$]{6,}$", password):
        return password
    else:
        return None

password = ['ABd1234@1','abcdef', '3Fa1#', '2w3E*', '2We3345']
pass_ok = []
for p in password:
    pass_ok.append(pass_check(p))
print(pass_ok)

Output:

['ABd1234@1', None, None, None, None]

Note:

1. (?=...)

   Matches if ... matches next, but doesn't consume any of the string. This is called a positive lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it's followed by 'Asimov'.

   Note that if we use '!' instead of '=', we call it negative lookahead. For example, Isaac (?!Asimov) will match 'Isaac ' if it's not followed by 'Asimov'.

   import re
   result = re.match(r'Issac (?=Asimov)', 'Issac Asimov 1920-1992')
   if result:
       print(result.group(0))
   else:
       print('no match')
       
   result = re.match(r'Issac (?!Asimov)', 'Issac Asimov 1920-1992')
   if result:
       print(result.group(0))
   else:
       print('no match')
   

   
   

   Output:

   Issac 
   no match
   



2. \d
   Matches any decimal digit

import re

# . (dot) : matches any character except newline
# \w      : matches any word character i.e letters, alphanumeric, digits and underscore ('_')
# +       : matches one or more of the preceding character (greedy match)

patterns = [r'\w@\w', r'\w+@\w+', r'[\w.-]+@[\w.]+']

for p in patterns:
    match = re.search(p, 'ki.hong-777@yahoo.com')
    if match:
        print('pattern = %s, match.group() = %s' %(p, match.group()))
    else:
        print('No match')

Output:

pattern = \w@\w, match.group() = 7@y
pattern = \w+@\w+, match.group() = 777@yahoo
pattern = [\w.-]+@[\w.]+, match.group() = ki.hong-777@yahoo.com

Group capturing allows us to extract parts from the matching string. We can create groups using parentheses ().

Using the capturing capability, we can extract username and host name from the email address in the above example. To do this we need to add () around username and host name like this:

r'([\w.-]+)@([\w.]+)'

The code:

import re

# . (dot) : matches any character except newline
# \w      : matches any word character i.e letters, alphanumeric, digits and underscore ('_')
# +       : matches one or more of the preceding character (greedy match)

patterns = [r'([\w.-]+)@([\w.]+)']

for p in patterns:
    match = re.search(p, 'ki.hong-777@yahoo.com')
    if match:
        print('pattern = %s, match.group() = %s' %(p, match.group()))
        print('group(1) = %s, group(2) = %s' %(match.group(1),match.group(2)))
    else:
        print('No match')

We can also use group capturing with findall(). When group capturing is applied, findall() returns a list of tuples where tuples will contain the matching groups:

import re

# +     : matches one or more of the preceding character (greedy match)
# \d    : matches a single digit

pattern = r'(\d{3})-(\d+)-(\d\d\d\d)'
numbers = ['123-456-7890', '3210-6540-09870']

print('pattern = %s' %pattern)
for n in numbers:
    match = re.findall(pattern, n)
    if match:       
        for m in match:
            print('number = %s, m = %s' %(n,m))
    else:
        print('No match')

Output:

pattern = (\d{3})-(\d+)-(\d\d\d\d)
number = 123-456-7890, m = ('123', '456', '7890')
number = 3210-6540-09870, m = ('210', '6540', '0987')

We used three patterns to illustrate the difference. The \d{3} picks up 3 digits preceding '-', \d+ gets any umber of digits before '-', and \d\d\d\d takes 4 digits as specified.

Write a function to find all the prime factors of a given integer. The function must return a list containing all the prime factors, sorted in ascending order.

import math
def prime_factors(n):
    pf = []
    while n % 2 == 0:
        pf.append(2)
        n = int(n/2)
    for i in range(3, int(n/2), 2):
        while n % i == 0:
            pf.append(i)
            n = n/i                   
    return pf

numbers = [2,5,12,34,135,741,5068,40356,362925,3628855,39916866,479001678]

for n in numbers:
    print('%s: %s' %(n, prime_factors(n)))

Output:

2: [2]
5: []
12: [2, 2]
34: [2]
135: [3, 3, 3, 5]
741: [3, 13, 19]
5068: [2, 2, 7, 181]
40356: [2, 2, 3, 3, 19, 59]
362925: [3, 3, 5, 5, 1613]
3628855: [5, 557, 1303]
39916866: [2, 3, 11, 604801]
479001678: [2, 3, 79833613]

Create a function that takes a string (IPv4 address in standard dot-decimal format) and returns True if the IP is valid or False if it's not.

# "1.2.3.4" => True
# "1.2.3" => False
# "1.2.3.4.5" => False
# "123.45.67.89" => True
# "123.456.78.90" => False
# "123.045.067.089" => False

def valid(ip):
  n4 = ip.split('.')
  if len(n4) != 4:
     return False
  for n in n4:
     if n[0] == '0' or int(n) > 255:
       return False

  return True

ips = ["1.2.3.4", "1.2.3", "1.2.3.4.5", "123.45.67.89", "123.456.78.90", "123.045.067.089"]

for ip in ips:
  print('%s => %s' %(ip,valid(ip))) 

Output:

1.2.3.4 => True
1.2.3 => False
1.2.3.4.5 => False
123.45.67.89 => True
123.456.78.90 => False
123.045.067.089 => False

Write a program to replace each string with an integer value in a given list of strings. The replacement integer value should be a sum of Ascii values of each character of the corresponding string.

countries  =['Denmark', 'Estonia', 'Finland', 'France', 'Germany', 'Greece', 'Hungary', 'Ireland', 'Italy']
ord_sum = []
for country in countries:
  ord_sum.append(sum([ord(c) for c in country]))
print(ord_sum)

Output:

[706, 723, 700, 591, 723, 587, 734, 703, 515]

Write a left and right rotations of a list. For example, rotation(l, 2) should convert a = [1,2,3,4,5] should print out [3,4,5,1,2], which is a left rotation. For right rotation, the 2nd parameter can be given as -2 instead of +2.

def rotation(a,n):
	return a[n:] + a[:n]
	
a = [1,2,3,4,5]

# left 
print(rotation(a,1))
print(rotation(a,2))

# right
print(rotation(a,-1))
print(rotation(a,-2))

Output:

[2, 3, 4, 5, 1]
[3, 4, 5, 1, 2]
[5, 1, 2, 3, 4]
[4, 5, 1, 2, 3]

For more complete (general) solution, please check Left rotation - list shift

The keyword here is reference!

We copy to make a duplicate data. But this copy function only makes the copy of reference that means if we change the new object then this will change the original as well.

So, we may want to use deep copy so that we can create a new object with original one and can modify the newly created object separately.

Note that the deep copy makes execution of the program slower due to making certain copies for each object that is been called.

Picture credit: trying to do a shallow copy on list in python

When a child element of a list changes, the element of the copied list also changes. To make the original and copied ones remain independent, we can use copy.deepcopy().

import copy

x = [1,2,[1,2]]
y = x
print('x=%s, y=%s' %(x,y))
x[2][0] = '?'
print('x=%s, y=%s' %(x,y))

x2 = [1,2,[1,2]]
y2 = copy.copy(x2)
print('x2=%s, y2=%s' %(x2,y2))
x2[2][0] = '?'
print('x2=%s, y2=%s' %(x2,y2))
                  
x3 = [1,2,[1,2]]
y3 = copy.deepcopy(x3)
print('x3=%s, y3=%s' %(x3,y3))
x3[2][0] = '?'
print('x3=%s, y3=%s' %(x3,y3))

Output:

x=[1, 2, [1, 2]], y=[1, 2, [1, 2]]
x=[1, 2, ['?', 2]], y=[1, 2, ['?', 2]]
x2=[1, 2, [1, 2]], y2=[1, 2, [1, 2]]
x2=[1, 2, ['?', 2]], y2=[1, 2, ['?', 2]]
x3=[1, 2, [1, 2]], y3=[1, 2, [1, 2]]
x3=[1, 2, ['?', 2]], y3=[1, 2, [1, 2]]

Note that for the deepcopied y3[2][0]=1, not affected from the x3[2][0]='?'
Note also that y = x is the same as y = copy.copy(x).

Another example: shallow copy vs deep copy.

def binary(n):
    b = []
    while n > 0:
        b.append(str(n % 2))
        n = n // 2
    return b[::-1]    

n = 21
bb = binary(n)
print(bb)
print(''.join(bb))

Output:

['1', '0', '1', '0', '1']
10101

Or we can just use bin():

n = 21
print(bin(n))
print(bin(n)[2:])

Output:

0b10101
10101

Binary => int:

int(0b1111) # =>  15
int(0B1111) # =>  15

Write a code to create a directory with name "Temporary" and create a file "myfile.txt" in the new directory "Temporary".

import os
filename = os.path.expanduser('~/Documents/TEST/Temporary/myfile.txt')
dirname = os.path.dirname(filename)

if not os.path.exists(dirname):
    os.makedirs(dirname)
with open(filename,'w') as f:
    f.write("test")

List all the files from a given path and verify the file "myfile.txt" file exists or not? If the file does not exist create it.

import os
path = os.path.expanduser('~/Documents/TEST/Temporary')
fname = "myfile.txt"
print(path)
for root, dirnames, filenames in os.walk(path):
    files = [f for f in filenames]
if not fname in files:
    with open(os.path.join(path, fname), 'w') as f:
        f.write("test...")

Create a hello.py file and run this program by invoking it in another python file.

# hello.py
print("hello, python!")
  
def hello_1():
   print('hello-1')

def hello_2():
   print('hello-2')

# call_hello.py
import hello
  
hello.hello_1()
hello.hello_2()

We can run it "python call_hello.py", and the output:

hello, python!
hello-1
hello-2

import operator

ips = ['5.4.3.2', '1.2.3.4', '5.4.1.2', '1.2.4.5']

# [['5', '4', '3', '2'], [], ...]
splitted = [(ip.split('.')) for ip in ips]

# [[['1', '2', '3', '4'], [], ...]]
sorted_splitted = sorted(splitted, key=operator.itemgetter(0,1,2,3), reverse=False)

# inserting dot('.')s
sorted_ips = ['.'.join(s) for s in sorted_splitted]
    
print('%s => %s' %(ips,sorted_ips))

Output:

['5.4.3.2', '1.2.3.4', '5.4.1.2', '1.2.4.5'] => ['1.2.3.4', '1.2.4.5', '5.4.1.2', '5.4.3.2']

For a given text, we want to get the frequency of the words. The most frequent one at the top.

text="""
bands which have connected them with another, 
and to assume among the powers of the earth, 
the separate and equal station to which the Laws of Nature 
and of Nature's God entitle them, 
a decent respect to the opinions of mankind requires 
that they should declare the causes which impel them to the separation.
"""

First, we need to process the text excluding some characters:

words = text.replace('-',' ').replace(',',' ').replace('.',' ').replace('\n',' ').lower().split()
print(words)

The modified input text should look like this:

['bands', 'which', 'have', 'connected', 'them', 'with', 'another', 'and', 'to', 'assume', 'among', 'the', 'powers', 'of', 'the', 'earth', 'the', 'separate', 'and', 'equal', 'station', 'to', 'which', 'the', 'laws', 'of', 'nature', 'and', 'of', "nature's", 'god', 'entitle', 'them', 'a', 'decent', 'respect', 'to', 'the', 'opinions', 'of', 'mankind', 'requires', 'that', 'they', 'should', 'declare', 'the', 'causes', 'which', 'impel', 'them', 'to', 'the', 'separation']

Here are some of the implementations:

d = {}.fromkeys(words,0)
for w in words:
    d[w] += 1
s = sorted(d.items(), key=lambda x:x[1], reverse=True)
print(s)

import operator
d = {}
for w in words:
    d[w] = d.get(w,0)+1
s = sorted(d.items(), key=operator.itemgetter(1), reverse=True)
print(s)

import collections
d = collections.Counter(words) # descending
#d = sorted(collections.Counter(words), reverse=True) # ascending
print(d)

d = {}
for w in words:
    if w not in d:
        d[w] = 0
    d[w] += 1
mylist = [(k, d[k]) for k in sorted(d, key=d.get, reverse=True)]
print(mylist)

The outputs should look like this:

[('the', 7), ('to', 4), ('of', 4), ('which', 3), ('them', 3), ('and', 3), ('bands', 1), ('have', 1), ('connected', 1), ('with', 1), ('another', 1), ('assume', 1), ('among', 1), ('powers', 1), ('earth', 1), ('separate', 1), ('equal', 1), ('station', 1), ('laws', 1), ('nature', 1), ("nature's", 1), ('god', 1), ('entitle', 1), ('a', 1), ('decent', 1), ('respect', 1), ('opinions', 1), ('mankind', 1), ('requires', 1), ('that', 1), ('they', 1), ('should', 1), ('declare', 1), ('causes', 1), ('impel', 1), ('separation', 1)]

With a given query string, we want to make a dictionary for the parameters. Note that the url has two identical key, "userEmail", and we want to keep both of the emails.

url = "https://bogotobogo.com/?userId=1&userName=Bob&userEmail=foo@gmail.com&userEmail=bar@yahoo.com"

from collections import defaultdict
parm_dict = defaultdict(list)
par_list = url.split('?')[1].split('&')
print(par_list)
for p in par_list:
    k,v = p.split('=')
    parm_dict[k].append(v)
    
print(parm_dict)

Output:

['userId=1', 'userName=Bob', 'userEmail=foo@gmail.com', 'userEmail=bar@yahoo.com']
defaultdict(<class 'list'>, {'userId': ['1'], 'userName': ['Bob'], 'userEmail': ['foo@gmail.com', 'bar@yahoo.com']})

Print numbers in spiral pattern (clock-wise) for a given 2D array.

# clock-wise spiral

def spiral(re, ce, a):
    
    rs = 0; cs = 0
    
    # rs : starting row index
    # rd : ending row index
    # cs : starting column index
    # ce : ending column index
    
    while (rs < re and cs < ce):
        
        # printing the 1st row of the remaining rows
        for c in range(cs, ce):
            print(a[rs][c],end=" ")
        rs += 1
        
        # printing the last column of the remaining columns
        for r in range(rs, re):
            print(a[r][ce-1],end=" ")
        ce -= 1
        
        # printing the last row of the remaining rows
        if rs < re:
            for c in range(ce-1, cs-1, -1):
                print(a[re-1][c],end=" ")
            re -= 1        

        # printing the first column of the remaining columns
        if cs < ce:
            for r in range(re-1, rs-1, -1):
                print(a[r][c],end=" ")
            cs += 1   

if __name__ == '__main__':
   
    # Case 1 : 4 x 6 grid
    row = 4
    column = 6
    arr = [ [ 1,  2,  3,  4,  5, 6],
            [16, 17, 18, 19, 20, 7],
            [15, 24, 23, 22, 21, 8],
            [14, 13, 12, 11, 10, 9]]
    spiral(4, 6, arr)

    print()
    
    # Case 2 - 5 x 5 grid
    row = 5
    column = 5
    arr = [ [ 1,  2,  3,  4, 5],
            [16, 17, 18, 19, 6],
            [15, 24, 25, 20, 7],
            [14, 23, 22, 21, 8],
            [13, 12, 11, 10, 9]]
    spiral(5, 5, arr)

Output:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 

Print numbers in spiral pattern (coundter-clock-wise) for a given 2D array.

# counter-clock-wise spiral

def spiral(re, ce, a):
    
    rs = 0; cs = 0
    
    # rs : starting row index
    # rd : ending row index
    # cs : starting column index
    # ce : ending column index
    
    while (rs < re and cs < ce):
        
        # printing the 1st column of the remaining columns
        for r in range(rs, re):
            print(a[r][cs],end=" ")
        cs += 1
        
        # printing the last row of the remaining rows
        for c in range(cs, ce):
            print(a[re-1][c],end=" ")
        re -= 1
        
        # printing the last column of the remaining columns
        if cs < ce:
            for r in range(re-1, rs-1, -1):
                print(a[r][ce-1],end=" ")
            ce -= 1        

        # printing the first row of the remaining rows
        if rs < re:
            for c in range(ce-1, cs-1, -1):
                print(a[r][c],end=" ")
            rs += 1   

if __name__ == '__main__':
   
    # Case 1 : 4 x 6 grid
    row = 4
    column = 6
    arr = [ [1, 15, 13, 14, 13, 12],
            [2, 16, 23, 22, 21, 11],
            [3, 17, 18, 19, 20, 10],
            [4,  5,  6,  7,  8,  9]]
    spiral(4, 6, arr)

    print()
    
    # Case 2 - 5 x 5 grid
    row = 5
    column = 5
    arr = [ [1, 16, 15, 14, 13],
            [2, 17, 24, 23, 12],
            [3, 18, 25, 22, 11],
            [4, 19, 20, 21, 10],
            [5,  6,  7,  8, 9]]
    spiral(5, 5, arr)

Output:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 13 15 16 17 18 19 20 21 22 23 
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 

Print out a minimum integer (should be greater than 0) not in the input list of integers. The size of the list could be up to N = 100000.

N = 100000
def my_integer(a):   
    a = set(a)
    i = 1
    while (i <= N):
        if i not in a and i >= 1:
            return i
        i += 1

a = [1, 2, 7, 5, 4, 1, 3]
print(my_integer(a))

a = [1, 2, 3] 
print(my_integer(a))
 
a =[2]
print(my_integer(a))

a = [x for x in range(-5,5,2)]
print(my_integer(a))

import random

a = [random.randint(-N//2, N//2) for x in range(N//2)]
print(my_integer(a))

a = [random.randint(-2, 2) for x in range(10)]
print(my_integer(a))

Output:

6
4
1
2
4
3

def binary(n):
    b = []
    while n > 0:
        b.append(n%2)
        n = n // 2
    return b

def mxlength(a):
    count = 0
    mx = 0
    for index, e in enumerate(a):
        if e == 0:
            count += 1
            mx = max(mx, count)
        else:
            count = 0
    return mx
            
for x in range(20):
    b = binary(x)
    print('%s : b=%s, mx=%s' %(x, b, mxlength(b)))

Output:

0 : b=[], mx=0
1 : b=[1], mx=0
2 : b=[0, 1], mx=1
3 : b=[1, 1], mx=0
4 : b=[0, 0, 1], mx=2
5 : b=[1, 0, 1], mx=1
6 : b=[0, 1, 1], mx=1
7 : b=[1, 1, 1], mx=0
8 : b=[0, 0, 0, 1], mx=3
9 : b=[1, 0, 0, 1], mx=2
10 : b=[0, 1, 0, 1], mx=1
11 : b=[1, 1, 0, 1], mx=1
12 : b=[0, 0, 1, 1], mx=2
13 : b=[1, 0, 1, 1], mx=1
14 : b=[0, 1, 1, 1], mx=1
15 : b=[1, 1, 1, 1], mx=0
16 : b=[0, 0, 0, 0, 1], mx=4
17 : b=[1, 0, 0, 0, 1], mx=3
18 : b=[0, 1, 0, 0, 1], mx=2
19 : b=[1, 1, 0, 0, 1], mx=2

The difference from the previous case is that the 0-block should be surrounded with 1s. So, the number 16 = [1,0,0,0,0] should have max = 0 not 4 because it's missing '1' at the end.

def binary(n):
    b = []
    while n > 0:
        b.append(n%2)
        n = n // 2
    return b 

def mxlength(b):
    count = 0
    mx = 0
    total = 0
    for index, bit in enumerate(b):
        if bit == 0:  
            # exclude the 0-block with starting and ending with 0
            if index == 0 or index == len(b)-1:
                count = 0
                total = 0
            # for each bit, calculate the sum of the bits before and after the index
            else:               
                count += 1
                total = total + b[index-1]+b[index+1]   
            # legitimate 0-block's total should be 2
            if total == 2:
                mx = max(mx, count)
        else:
            count = 0
            total = 0
    return mx

for x in range(100):
    b = binary(x)
    print('%s : b=%s, mx=%s' %(x, b, mxlength(b)))

Output:

0 : b=[], mx=0
1 : b=[1], mx=0
2 : b=[0, 1], mx=0
3 : b=[1, 1], mx=0
4 : b=[0, 0, 1], mx=0
5 : b=[1, 0, 1], mx=1
6 : b=[0, 1, 1], mx=0
7 : b=[1, 1, 1], mx=0
8 : b=[0, 0, 0, 1], mx=0
9 : b=[1, 0, 0, 1], mx=2
10 : b=[0, 1, 0, 1], mx=1
11 : b=[1, 1, 0, 1], mx=1
12 : b=[0, 0, 1, 1], mx=0
13 : b=[1, 0, 1, 1], mx=1
14 : b=[0, 1, 1, 1], mx=0
15 : b=[1, 1, 1, 1], mx=0
16 : b=[0, 0, 0, 0, 1], mx=0
17 : b=[1, 0, 0, 0, 1], mx=3
18 : b=[0, 1, 0, 0, 1], mx=2
19 : b=[1, 1, 0, 0, 1], mx=2
20 : b=[0, 0, 1, 0, 1], mx=1
21 : b=[1, 0, 1, 0, 1], mx=1
22 : b=[0, 1, 1, 0, 1], mx=1
23 : b=[1, 1, 1, 0, 1], mx=1
24 : b=[0, 0, 0, 1, 1], mx=0
25 : b=[1, 0, 0, 1, 1], mx=2
26 : b=[0, 1, 0, 1, 1], mx=1
27 : b=[1, 1, 0, 1, 1], mx=1
28 : b=[0, 0, 1, 1, 1], mx=0
29 : b=[1, 0, 1, 1, 1], mx=1
30 : b=[0, 1, 1, 1, 1], mx=0
31 : b=[1, 1, 1, 1, 1], mx=0
32 : b=[0, 0, 0, 0, 0, 1], mx=0
33 : b=[1, 0, 0, 0, 0, 1], mx=4
34 : b=[0, 1, 0, 0, 0, 1], mx=3
35 : b=[1, 1, 0, 0, 0, 1], mx=3
36 : b=[0, 0, 1, 0, 0, 1], mx=2
37 : b=[1, 0, 1, 0, 0, 1], mx=2
38 : b=[0, 1, 1, 0, 0, 1], mx=2
39 : b=[1, 1, 1, 0, 0, 1], mx=2
40 : b=[0, 0, 0, 1, 0, 1], mx=1
41 : b=[1, 0, 0, 1, 0, 1], mx=2
42 : b=[0, 1, 0, 1, 0, 1], mx=1
43 : b=[1, 1, 0, 1, 0, 1], mx=1
44 : b=[0, 0, 1, 1, 0, 1], mx=1
45 : b=[1, 0, 1, 1, 0, 1], mx=1
46 : b=[0, 1, 1, 1, 0, 1], mx=1
47 : b=[1, 1, 1, 1, 0, 1], mx=1
48 : b=[0, 0, 0, 0, 1, 1], mx=0
49 : b=[1, 0, 0, 0, 1, 1], mx=3
50 : b=[0, 1, 0, 0, 1, 1], mx=2
51 : b=[1, 1, 0, 0, 1, 1], mx=2
52 : b=[0, 0, 1, 0, 1, 1], mx=1
53 : b=[1, 0, 1, 0, 1, 1], mx=1
54 : b=[0, 1, 1, 0, 1, 1], mx=1
55 : b=[1, 1, 1, 0, 1, 1], mx=1
56 : b=[0, 0, 0, 1, 1, 1], mx=0
57 : b=[1, 0, 0, 1, 1, 1], mx=2
58 : b=[0, 1, 0, 1, 1, 1], mx=1
59 : b=[1, 1, 0, 1, 1, 1], mx=1
60 : b=[0, 0, 1, 1, 1, 1], mx=0
61 : b=[1, 0, 1, 1, 1, 1], mx=1
62 : b=[0, 1, 1, 1, 1, 1], mx=0
63 : b=[1, 1, 1, 1, 1, 1], mx=0
64 : b=[0, 0, 0, 0, 0, 0, 1], mx=0
65 : b=[1, 0, 0, 0, 0, 0, 1], mx=5
66 : b=[0, 1, 0, 0, 0, 0, 1], mx=4
67 : b=[1, 1, 0, 0, 0, 0, 1], mx=4
68 : b=[0, 0, 1, 0, 0, 0, 1], mx=3
69 : b=[1, 0, 1, 0, 0, 0, 1], mx=3
70 : b=[0, 1, 1, 0, 0, 0, 1], mx=3
71 : b=[1, 1, 1, 0, 0, 0, 1], mx=3
72 : b=[0, 0, 0, 1, 0, 0, 1], mx=2
73 : b=[1, 0, 0, 1, 0, 0, 1], mx=2
74 : b=[0, 1, 0, 1, 0, 0, 1], mx=2
75 : b=[1, 1, 0, 1, 0, 0, 1], mx=2
76 : b=[0, 0, 1, 1, 0, 0, 1], mx=2
77 : b=[1, 0, 1, 1, 0, 0, 1], mx=2
78 : b=[0, 1, 1, 1, 0, 0, 1], mx=2
79 : b=[1, 1, 1, 1, 0, 0, 1], mx=2
80 : b=[0, 0, 0, 0, 1, 0, 1], mx=1
81 : b=[1, 0, 0, 0, 1, 0, 1], mx=3
82 : b=[0, 1, 0, 0, 1, 0, 1], mx=2
83 : b=[1, 1, 0, 0, 1, 0, 1], mx=2
84 : b=[0, 0, 1, 0, 1, 0, 1], mx=1
85 : b=[1, 0, 1, 0, 1, 0, 1], mx=1
86 : b=[0, 1, 1, 0, 1, 0, 1], mx=1
87 : b=[1, 1, 1, 0, 1, 0, 1], mx=1
88 : b=[0, 0, 0, 1, 1, 0, 1], mx=1
89 : b=[1, 0, 0, 1, 1, 0, 1], mx=2
90 : b=[0, 1, 0, 1, 1, 0, 1], mx=1
91 : b=[1, 1, 0, 1, 1, 0, 1], mx=1
92 : b=[0, 0, 1, 1, 1, 0, 1], mx=1
93 : b=[1, 0, 1, 1, 1, 0, 1], mx=1
94 : b=[0, 1, 1, 1, 1, 0, 1], mx=1
95 : b=[1, 1, 1, 1, 1, 0, 1], mx=1
96 : b=[0, 0, 0, 0, 0, 1, 1], mx=0
97 : b=[1, 0, 0, 0, 0, 1, 1], mx=4
98 : b=[0, 1, 0, 0, 0, 1, 1], mx=3
99 : b=[1, 1, 0, 0, 0, 1, 1], mx=3

Note that the returned list from the function (binary()) should be reversed for the bits to represent the real binary value. In our case, however, it gives us the right answer.

A list consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.

Let's try to find the missing element.

def missing(a):
    exists = [False]*(len(A)+1)
    for item in a:
        exists[item-1] = True
    for index, e in enumerate(exists):
        if e == False:
            return index+1

a = [1,2,3,5,6]
print(missing(a))

Output:

4

A non-empty list a consisting of N integers is given. The list contains an odd number of elements, and each element of the list can be paired with another element that has the same value, except for one element that is left unpaired.

For example, given a = [1,2,3,1,2], the output should be '3'.

The following solution gives us the right answer, but it's O(n^2):

def unpaired(a):
    for e in a:
        if a.count(e) % 2 == 1:
            return e
    return None

a = [1,2,3,1,2]
unpaired(a)

Another try which looks ok - (O(N logN)) and traverse(O(N)):

def unpaired(a):
    d = dict.fromkeys(a,0)
    print(d)
    for x in a:
        d[x] += 1
    print(d)
    
    odd = []
    for k,v in d.items():
        if v % 2 == 1:
            odd.append(k)
            return k
    
    return k
            
a = [9,3,9,3,9,7,9]
print(unpaired(a))

Here we have a final solution: sort the list first ( O(N logN) ), and traverse( O(N) ) the elements with step 2:

def unpaired(a):
    if len(a) == 1:
        return a[0]
    
    a = sorted(a)
    for i in range(0,len(a),2):
        if i+1 == len(a):
            return a[i]
        if a[i] != a[i+1]:
            return a[i]

a = [1,2,3,5,6,10,1,2,6,3,5]
print(unpaired(a))

import random
N = 1000000
a = [random.randint(1,N) for x in range(N//2)]
print(unpaired(a))

Output:

10
2

1. Python Coding Questions I
2. Python Coding Questions II
3. Python Coding Questions III
4. Python Coding Questions IV
5. Python Coding Questions V
6. Python Coding Questions VI
7. Python Coding Questions VII
8. Python Coding Questions VIII
9. Python Coding Questions IX
10. Python Coding Questions X

List of codes Q & A

1. Merging two sorted list
2. Get word frequency - initializing dictionary
3. Initializing dictionary with list
4. map, filter, and reduce
5. Write a function f() - yield
6. What is __init__.py?
7. Build a string with the numbers from 0 to 100, "0123456789101112..."
8. Basic file processing: Printing contents of a file - "with open"
9. How can we get home directory using '~' in Python?
10. The usage of os.path.dirname() & os.path.basename() - os.path
11. Default Libraries
12. range vs xrange
13. Iterators
14. Generators
15. Manipulating functions as first-class objects
16. docstrings vs comments
17. using lambdda
18. classmethod vs staticmethod
19. Making a list with unique element from a list with duplicate elements
20. What is map?
21. What is filter and reduce?
22. *args and **kwargs
23. mutable vs immutable
24. Difference between remove, del and pop on lists
25. Join with new line
26. Hamming distance
27. Floor operation on integers
28. Fetching every other item in the list
29. Python type() - function
30. Dictionary Comprehension
31. Sum
32. Truncating division
33. Python 2 vs Python 3
34. len(set)
35. Print a list of file in a directory
36. Count occurrence of a character in a Python string
37. Make a prime number list from (1,100)
38. Reversing a string - Recursive
39. Reversing a string - Iterative
40. Reverse a number
41. Output?
42. Merging overlapped range
43. Conditional expressions (ternary operator)
44. Packing Unpacking
45. Function args
46. Unpacking args
47. Finding the 1st revision with a bug
48. Which one has higher precedence in Python? - NOT, AND , OR
49. Decorator(@) - with dollar sign($)
50. Multi-line coding
51. Recursive binary search
52. Iterative binary search
53. Pass by reference
54. Simple calculator
55. iterator class that returns network interfaces
56. Converting domain to ip
57. How to count the number of instances
58. Python profilers - cProfile
59. Calling a base class method from a child class that overrides it
60. How do we find the current module name?
61. Why did changing list 'newL' also change list 'L'?
62. Constructing dictionary - {key:[]}
63. Colon separated sequence
64. Converting binary to integer
65. 9+99+999+9999+...
66. Calculating balance
67. Regular expression - findall
68. Chickens and pigs
69. Highest possible product
70. Implement a queue with a limited size
71. Copy an object
72. Filter
73. Products
74. Pickle
75. Overlapped Rectangles
76. __dict__
77. Fibonacci I - iterative, recursive, and via generator
78. Fibonacci II - which method?
79. Fibonacci III - find last two digits of Nth Fibonacci number
80. Write a Stack class returning Max item at const time A
81. Write a Stack class returning Max item at const time B
82. Finding duplicate integers from a list - 1
83. Finding duplicate integers from a list - 2
84. Finding duplicate integers from a list - 3
85. Reversing words 1
86. Parenthesis, a lot of them
87. Palindrome / Permutations
88. Constructing new string after removing white spaces
89. Removing duplicate list items
90. Dictionary exercise
91. printing numbers in Z-shape
92. Factorial
93. lambda
94. lambda with map/filter/reduce
95. Number of integer pairs whose difference is K
96. iterator vs generator
97. Recursive printing files in a given directory
98. Bubble sort
99. What is GIL (Global Interpreter Lock)?
100. Word count using collections
101. Pig Latin
102. List of anagrams from a list of words
103. lamda with map, filer and reduce functions
104. Write a code sending an email using gmail
105. histogram 1 : the frequency of characters
106. histogram 2 : the frequency of ip-address
107. Creating a dictionary using tuples
108. Getting the index from a list
109. Looping through two lists side by side
110. Dictionary sort with two keys : primary / secondary keys
111. Writing a file downloaded from the web
112. Sorting csv data
113. Reading json file
114. Sorting class objects
115. Parsing Brackets
116. Printing full path
117. str() vs repr()
118. Missing integer from a sequence
119. Polymorphism
120. Product of every integer except the integer at that index
121. What are accessors, mutators, and @property?
122. N-th to last element in a linked list
123. Implementing linked list
124. Removing duplicate element from a list
125. List comprehension
126. .py vs .pyc
127. Binary Tree
128. Print 'c' N-times without a loop
129. Quicksort
130. Dictionary of list
131. Creating r x c matrix
132. Transpose of a matrix
133. str.isalpha() & str.isdigit()
134. Regular expression
135. What is Hashable? Immutable?
136. Convert a list to a string
137. Convert a list to a dictionary
138. List - append vs extend vs concatenate
139. Use sorted(list) to keep the original list
140. list.count()
141. zip(list,list) - join elements of two lists
142. zip(list,list) - weighted average with two lists
143. Intersection of two lists
144. Dictionary sort by value
145. Counting the number of characters of a file as One-Liner
146. Find Armstrong numbers from 100-999
147. Find GCF (Greatest common divisor)
148. Find LCM (Least common multiple)
149. Draws 5 cards from a shuffled deck
150. Dictionary order by value or by key
151. Regular expression - re.split()
152. Regular expression : re.match() vs. re.search()
153. Regular expression : re.match() - password check
154. Regular expression : re.search() - group capturing
155. Regular expression : re.findall() - group capturin
156. Prime factors : n = products of prime numbers
157. Valid IPv4 address
158. Sum of strings
159. List rotation - left/right
160. shallow/deep copy
161. Converting integer to binary number
162. Creating a directory and a file
163. Creating a file if not exists
164. Invoking a python file from another
165. Sorting IP addresses
166. Word Frequency
167. Printing spiral pattern from a 2D array - I. Clock-wise
168. Printing spiral pattern from a 2D array - II. Counter-Clock-wise
169. Find a minimum integer not in the input list
170. I. Find longest sequence of zeros in binary representation of an integer
171. II. Find longest sequence of zeros in binary representation of an integer - should be surrounded with 1
172. Find a missing element from a list of integers
173. Find an unpaired element from a list of integers
174. Prefix sum : Passing cars
175. Prefix sum : count the number of integers divisible by k in range [A,B]
176. Can make a triangle?
177. Dominant element of a list
178. Minimum perimeter
179. MinAbsSumOfTwo
180. Ceiling - Jump Frog
181. Brackets - Nested parentheses
182. Brackets - Nested parentheses of multiple types
183. Left rotation - list shift
184. MaxProfit
185. Stack - Fish
186. Stack - Stonewall
187. Factors or Divisors
188. String replace in files 1
189. String replace in files 2
190. Using list as the default_factory for defaultdict
191. Leap year
192. Capitalize
193. Log Parsing
194. Getting status_code for a site
195. 2D-Array - Max hourglass sum
196. New Year Chaos - list
197. List (array) manipulation - list
198. Hash Tables: Ransom Note
199. Count Triplets with geometric progression
200. Strings: Check if two strings are anagrams
201. Strings: Making Anagrams
202. Strings: Alternating Characters
203. Special (substring) Palindrome
204. String with the same frequency of characters
205. Common Child
206. Fraudulent Activity Notifications
207. Maximum number of toys
208. Min Max Riddle
209. Poisonous Plants with Pesticides
210. Common elements of 2 lists - Complexity
211. Get execution time using decorator(@)
212. Conver a string to lower case and split using decorator(@)
213. Python assignment and memory location
214. shallow copy vs deep copy for compound objects (such as a list)
215. Generator with Fibonacci
216. Iterator with list
217. Second smallest element of a list
218. *args, **kargs, and positional args
219. Write a function, fn('x','y',3) that returns ['x1', 'y1', 'x2', 'y2', 'x3', 'y3']
220. sublist or not
221. any(), all()
222. Flattening a list
223. Select an element from a list
224. Circularly identical lists
225. Difference between two lists
226. Reverse a list
227. Split a list with a step
228. Break a list and make chunks of size n
229. Remove duplicate consecutive elements from a list
230. Combination of elements from two lists
231. Adding a sublist
232. Replace the first occurence of a value
233. Sort the values of the first list using the second list
234. Transpose of a matrix (nested list)
235. Binary Gap
236. Powerset
237. Round Robin
238. Fixed-length chunks or blocks
239. Accumulate
240. Dropwhile
241. Groupby
242. Simple product
243. Simple permutation
244. starmap(fn, iterable)
245. zip_longest(*iterables, fillvalue=None)
246. What is the correct way to write a doctest?
247. enumerate(iterable, start=0)
248. collections.defaultdict - grouping a sequence of key-value pairs into a dictionary of lists
249. What is the purpose of the 'self' keyword when defining or calling instance methods?
250. collections.namedtuple(typename, field_names, *, rename=False, defaults=None, module=None)
251. zipped
252. What is key difference between a set and a list?
253. What does a class's init() method do?
254. Class methods
255. Permutations and combinations of ['A','B','C']
256. Sort list of dictionaries by values
257. Return a list of unique words
258. hashlib
259. encode('utf-8')
260. Reading in CSV file
261. Count capital letters in a file
262. is vs ==
263. Create a matrix : [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
264. Binary to integer and check if it's the power of 2
265. urllib.request.urlopen() and requests
266. Game statistics
267. Chess - pawn race
268. Decoding a string
269. Determinant of a matrix - using numpy.linalg.det()
270. Revenue from shoe sales - using collections.Counter()
271. Rangoli
272. Unique characters
273. Valid UID
274. Permutations of a string in lexicographic sorted order
275. Nested list
276. Consecutive digit count
277. Find a number that occurs only once
278. Sorting a two-dimensional array
279. Reverse a string
280. Generate random odd numbers in a range
281. Shallow vs Deep copy
282. Transpose matrix
283. Are Arguments in Python Passed by Value or by Reference?
284. re: Is a string alphanumeric?
285. reversed()
286. Caesar's cipher, or shift cipher, Caesar's code, or Caesar shift
287. Every other words
288. re: How can we check if an email address is valid or not?
289. re: How to capture temperatures of a text
290. re.split(): How to split a text.
291. How can we merge two dictionaries?
292. How can we combine two dictionaries?
293. What is the difference between a generator and a list?
294. Pairs of a given array A whose sum value is equal to a target value N
295. Adding two integers without plus
296. isinstance() vs type()
297. What is a decorator?
298. In Python slicing, what does my_list[-3:2:-2] slice do?
299. Revisit sorting dict - counting chars in a text file
300. re: Transforming a date format using re.sub
301. How to replace the newlines in csv file with tabs?
302. pandas.merge
303. How to remove duplicate charaters from a string?
304. Implement a class called ComplexNumber
305. Find a word frequency
306. Get the top 3 most frequent characters of a string
307. Just seen and ever seen
308. Capitalizing the full name
309. Counting Consequitive Characters
310. Calculate Product of a List of Integers Provided using input()
311. How many times a substring appears in a string
312. Hello, first_name last_name
313. String validators
314. Finding indices that a char occurs in a list
315. itertools combinations