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Python puzzles to improve yourself as a junior coder

Development | Game Development
Description
Been a python coder for some time, I do find programming puzzles as a very interesting way to keep my brain in shape and also what's most important, to improve my coding skills. Having contact and friendship with veteran python developers, I have come to the conclusion that critical thinking and algorithmic solving skills are a must for a professional software programmer.
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Puzzle 1
Define a function which takes an integer number as input and print the "It is an even number" if the number is even, otherwise print "It is an odd number".

Solution

Based on simple arithmetic, an even number returns 0 for the remainder when divided by 2. For example, 4/2 =2 and the remainder is 0. With such logic, we can easily find out if the remainder is 0 or not by making usage of the builtin python operator % .

def check_num(num):
if num % 2 == 0:
print("It's an even number.") else:
print("It's an odd number.")
view rawpuzzle1.py hosted with ❤ by GitHub



Puzzle 2

Find the minimum of an array without using builtin, then find the maximum.

Solution
First, we setup the first element of the array as the minimum. Then, with the help of a for loop, we find the next minimum, until we have looped the whole array.

def 
find_min(array=[]):
minimo = array[0]
for el in array[1:]:
if minimo > 
el:
minimo = el
return minimo
view rawpuzzle2_solution.py hosted with ❤ by GitHub

We do the opposite for finding the maximum.


Puzzle 3

Define a function which can print a dictionary where the keys are numbers between 1 and 3 (both included) and the values are the square of the keys.

Solution

First we define the keys of the dictionary inside a list. Then with the help of a for loop, we assign the key value pairs to the dictionary.

def gen_dict():
d = {}
keys = [1, 2, 3]
for el in
keys:
d[el] = el * el
return


Puzzle 4

Write a program to generate and print another tuple whose values are even numbers in the given tuple (1, 2, 3, 4, 5, 6, 7, 8, 9, 10).

Solution

The trick in here is to store the even numbers inside a list object and once finished the population, to convert it into a tuple object.


def 
find_min(array=[]):
minimo = array[0]
for el in array[1:]:
if minimo > 
el:
minimo = el
return minimo

view rawpuzzle4.py hosted with ❤ by GitHub

Puzzle 5

Write a program which accepts a string as input to print "Yes" if the string is "yes" or "YES" or "Yes", otherwise print "No".

Solution
The key in here is to store the strings "yes", "Yes" and "YES" inside a list. Then do a check for the input within the list.

def check_string():
check_list = ["yes", "Yes", "YES"]
sq = raw_input("Enter the sequence:")
if sq in check_list:
print("Yes")
else:
print("No")
view rawpuzzle5.py hosted with ❤ by GitHub

Puzzle 6

Write a function to compute 5/0 and use try/except to catch the exceptions.

Solution

def doesnt_throw():
try:
5/0
except ZeroDivisionError:
print("Division by zero.")
except Exception, err:
print("Caught an exception.")
finally:
print("Final block for cleanup.")

view rawpuzzle6.py hosted with ❤ by GitHub


Puzzle 7

Make use of the assert statement to verify that every number in the list [2, 4, 6, 8] is even.

Solution

def check_even():
l = [2, 4, 6, 8]
for el in l:
assert(el%2 == 0)
view rawpuzzle7.py hosted with ❤ by GitHub

Puzzle 8

Add two numbers without using the addition operator.

Solution

A direct idea is to make usage of the subtract operator. That would work only for the positive numbers, so it's not a solution. The following is a better approach.


def add_nums_no_plus(a, b):

while (a > 0):
a -= 1
b += 1

while a < 0:
b -=1
a += 1
return b
view rawpuzzle8.py hosted with ❤ by GitHub

Puzzle 9

Sort an array without using the builtin method. [13, 14, 11, 2, 1] --> [1, 2, 11, 13, 14]

Solution

My idea of solving this puzzle is to create a fresh empty list and then populate it with each minimum of the given array.



def sort_array(array):
# [13, 14, 11, 2, 1] --> [1, 2, 11, 13, 14]
sorted_list = []
while len(array) > 0:
minimal = find_min(array)
sorted_list.append(minimal)
array.remove(minimal)
return sorted_list
view rawpuzzle9.py hosted with ❤ by GitHub
As you can see from the solution provided above, we make use of the find_min from the second puzzle.


Puzzle 10

Write a program which accepts a sentence and calculates the number of uppercase letters and lowercase letters. Suppose the following input is provided to the program:
Hello World!
Then the output should be:
UPPER CASE 2
LOWER CASE 9

Solution

def uppercase_lowercase():
d = {'UPPER_CASE':[], 'LOWER_CASE':[]}
sequence = raw_input("Type here:")
for c in sequence:
if c.isupper():
d['UPPER_CASE'].append(c)
elif c.islower():
d['LOWER_CASE'].append(c)

print('UPPER CASE %d' % len(d['UPPER_CASE']))
print('LOWER CASE %s' % len(d['LOWER_CASE']))
view rawpuzzle10.py hosted with ❤ by GitHub



Puzzle 11

Write a program to compute the factorial of a given number.

Solution

The builtin range function becomes useful in our case. We compute the list of factors with the help of it, and then we generate the factorial like shown in the following piece of code.

def factorial(number):
result = 1
factors = range(number+1)[1:]
for factor in factors:
result *= factor

return result
view rawpuzzle11.py hosted with ❤ by GitHub


Although the above algorithm works fine, the following solution is a more professional and elegant one.

def fact(num):
if num == 0:
return 1
return num * fact(num-1)
view rawpuzzle11_elegant.py hosted with ❤ by GitHub

Puzzle 12

Write a program that computes a value of a + aa + aaa + aaaa with a given digit as the value of a. Suppose the following input is supplied to the program:
9
Then, the output should be:
11106

Solution

def compute_expression():
a = raw_input("Enter digit:")
num1 = int("%s" % a)
num2 = int("%s%s" % (a, a))
num3 = int("%s%s%s" % (a, a, a))
num4 = int("%s%s%s%s" % (a, a, a, a))
s = num1 + num2 + num3 + num4
return s
view rawpuzzle12.py hosted with ❤ by GitHub

Puzzle 13

Write a program which will find all the numbers which are divisible by 7 but are not a multiple of 5, between 2000 and 3200. The numbers should be printed in a comma-separated sequence on a single line.

Solution

def divisible7():
nums = []
for el in range(2001, 3200):
if el % 7 == 0 and el % 5 != 0:
nums.append(el)
line = ','.join(str(num) for num in nums)

print(line)
view rawpuzzle13.py hosted with ❤ by GitHub

Puzzle 14

Write a program to generate all the sentences where subject is in ["I", "You"], verb is in ["Play", "Love"] and the object is in ["Hockey", "Football"].

Solution

The solution to this puzzle involves multiple for loops.

def all_sentences():
sentence = ""
subjects = ["I", "You"]
verbs = ["Play", "Love"]
objects = ["Hockey", "Football"]
for subject in subjects:
for verb in verbs:
for obj in objects:
sentence = " ".join([subject, verb, obj])
print(sentence)
sentence = ""
view rawpuzzle14.py hosted with ❤ by GitHub


Puzzle 15

Write a program which accepts a sequence of comma separated values from the console and generates a list and a tuple which contains every number. Suppose the following input is supplied to the program:

34, 67, 55, 33, 12, 98

Then the output should be:

[34, 67, 55, 33, 12, 98]
(34, 67, 55, 33, 12, 98)

Solution

def sequence_convert():
sequence = raw_input()
l = sequence.split(',')
t = tuple(l)
print(l)
print(t)
view rawpuzzle15.py hosted with ❤ by GitHub

Puzzle 16

Define a function that can accept two strings as input and print the string with maximum length on the console. If two strings have the same length, then the function should print both the strings line by line.

Solution

The if, else, elif conditionals can be easily implemented for the solution of this puzzle.

def longer_string(s1, s2):
first_length = len(s1)
second_length = len(s2)
if first_length - second_length > 0:
print(s1)
elif first_length - second_length == 0:
print(s1)
print(s2)
else:
print(s2)
view rawpuzzle16.py hosted with ❤ by GitHub



Puzzle 17

Write a program to filter even numbers in a list using the filter function. The list is [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].

Solution

For those of you who are not aware of the filter function, it is a builtin utility which takes for arguments a function and an iterable with the main purpose of filtering any element which proves True to the provided function.

def filter_list():
l = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
even_nums = filter(lambda x:x%2==0, l)
print(even_nums)
view rawpuzzle17.py hosted with ❤ by GitHub

Puzzle 18

When squirrels get together for a party, they like to have cigars. A squirrel party is successful when the number of cigars is between 40 and 60, inclusive. Unless it is the weekend, in which case there is no upper bound on the number of cigars. Return True if the party with the given values is successful, or False otherwise.

cigar_party(30, False)->False
cigar_party(50, False)->True
cigar_party(70, True)->True
cigar_party(70, False)->False

Solution

def cigar_party(cigars, is_weekend):
if cigars >= 40:
if is_weekend:
return True
else:
return cigars <= 60

return False
view rawpuzzle18.py hosted with ❤ by GitHub


Puzzle 19

Given two ints a, and b, return their sum. However, sums in the range 10..19 inclusive, are forbidden, so in that case just return 20.

Solution

def sorta_sum(a, b):
total = a + b
if total in range(10, 20):
return 20
return total
view rawpuzzle19.py hosted with ❤ by GitHub



Puzzle 20

The number 6 is truly a great number. Given two int values, a and b, return True if either one is 6. Or if their sum or difference is 6.

# love6(6, 4)-->True
# love6(4, 5)-->False
# love6(1, 5)-->True

Solution

The abs builtin functionality becomes useful in such case.

def love6(a, b):
if a == 6 or b == 6:
return True
if abs(a-b) == 6 or (a+b) == 6:
return True
return False
view rawpuzzle20.py hosted with ❤ by GitHub



Puzzle 21

Given three int values, a b c, return their sum. However, if one of the values is 13, then it does not count towards the sum and the values to its right do not count. So for example, if b is 13 then both b and c do not count.

Solution

def lucky_sum(a, b, c):
if a == 13:
return 0
elif b == 13:
return a
elif c == 13:
return a+b
else:
return a + b + c
view rawpuzzle21.py hosted with ❤ by GitHub


Puzzle 22

For this problem, we'll round an int value up to the next multiple of 10 if its rightmost digit is 5 or more, so 15 rounds up to 20. Alternately, round down to the previous multiple of 10 if its rightmost digit is less than 5, so 12 rounds down to 10. Given 3 ints, a b c, return the sum of their rounded values. To avoid code repetition, write a separate helper "def round10(num):" and call it 3 times.

Solution

def round10(num):
if num < 5:
return 0
if num % 10 == 0:
return num
if num % 10 >= 5:
return num + (10 - (num % 10))
if num % 10 < 5:
return num - (num % 10)


def round_sum(a, b, c):
return round10(a) + round10(b) + round10(c)
view rawpuzzle22.py hosted with ❤ by GitHub

Puzzle 23

Given 3 int values, a b c, return their sum. However, if any of the values is a teen -- in the range 13..19 inclusive-- then that value counts as 0, except 15 and 16 do not count as teens. Write a separate helper "def fix_teen(n):" that takes n as an int value and returns that value fixed for the teen rule.

# no_teen_sum(1, 2, 3)-->6
# no_teen_sum(2, 13, 1)-->3
# no_teen_sum(2, 1, 14-->3

Solution

def fix_teen(n):
if n in range(11, 20):
if n not in (15, 16):
return n
return 0
return n

def no_teen_sum(a, b, c):
return fix_teen(a) + fix_teen(b) + fix_teen(c)
view rawpuzzle23.py hosted with ❤ by GitHub

Puzzle 24

Given 3 int values, a b c, return their sum. However, if one of the values is the same as another of the values, it does not count toward the sum.

# lone_sum(1, 2, 3)->6
# lone_sum(3, 2, 3)->2
# lone_sum(3, 3, 3)->0

Solution

def lone_sum(a, b, c):
if a==b and a==c:
return 0
elif a==b and a!=c:
return c
elif a==c and a!=b:
return b
elif c==b and c!=a:
return a
else:
return a + b + c
view rawpuzzle24.py hosted with ❤ by GitHub


Puzzle 25

We want to make a package of goal kilos of chocolate. We have small bars (1 kilo each) and big bars (5 kilos each).Return the number of small bars to use, assuming we always use big bars before small bars. Return -1 if it can't be done.

# make_chocolate(4, 1, 9)-->4
# make_chocolate(4, 1, 10)-->-1
# make_chocolate(4, 1, 7)--> 2
# make_chocolate(6, 1, 10)-->5

Solution

def make_chocolate(small, big, goal):
# we use the big bars first
if goal/5 <= big:
if goal % 5 <= small:
return goal % 5
else:
return -1
if goal/5 >= big:
if (goal/5 - big) * 5 + (goal%5) <= small:
return (goal/5 -big)*5 + goal % 5
else:
return -1
view rawpuzzle25.py hosted with ❤ by GitHub


Puzzle 26

We want to make a row of bricks that is goal inches long. We have a number of small bricks(1 inch each) and big bricks(5 inches each). Return True if it is possible to make the goal by choosing from the given bricks.


# make_bricks(3, 1, 8)-->True
# make_bricks(3, 1, 9)-->False
# make_bricks(3, 2, 10)--True

Solution

def make_bricks(small, big, goal):
if goal < 5:
return goal <= small

if goal/5 >= big:
return ((goal/5 - big)*5 + (goal % 5)) <= small

if goal/5 <= big:
return (goal % 5) <= small
view rawpuzzle26.py hosted with ❤ by GitHub


Puzzle 27

You and your date are trying to get a table at a restaurant. The parameter "you" is the stylishness of your clothes, in the range 0..10, and "date" is the stylishness of your date's clothes.The result getting the table is encoded as int value with 0=no, 1=maybe, 2=yes. If either of you is very stylish, 8 or more, then the result is 2 (yes). With the exception that either of you has style of 2 or less, then the result is 0(no).
Otherwise the result is 1 (maybe).

# date_fashion(5, 10)->2
# date_fashion(5, 2)->0
# date_fashion(5, 5)->1

Solution

def date_fashion(you, date):
if you <=2 or date <=2:
return 0
if you >= 8 or date >= 8:
return 2
else:
return 1
view rawpuzzle27.py hosted with ❤ by GitHub

Puzzle 28

The squirrels in Palo Alto spend most of the day playing.In particular, they playif the temperature is between 60 and 90. Unless it is summer, then the upper limit is 100 instead of 90. Given an int temperature and a boolean is_summer, return True if the squirrels play and False otherwise.

# squirrel_play(70, False)->True
# squirrel_play(95, False)->False
# squirrel_play(95, True)->True

Solution

def squirrel_play(temp, is_summer):
if temp>=60:
if is_summer:
return temp<=100
else:
return temp<=90
return False
view rawpuzzle28.py hosted with ❤ by GitHub


Puzzle 29

Given a day of the week encoded as 0=Sun, 1=Mon, 2=Tue,...6=Sat, and a boolean indicating if we are
on a vacation, return a string of the form "7:00" indicating when the alarm clock should ring.Weekdays,
the arlam should be "7:00" and on the weekend it should be "10:00".Unless we are on vacation--then on weekdays it should be "10:00" and weekends it should be "off".


# alarm_clock(1, False)->'7:00'
# alarm_clock(5, False)->'7:00'
# alarm_clock(0, False)->'10:00'

Solution

def alarm_clock(day, vacation):
first_alarm = "7:00"
second_alarm = "10:00"
if day in range(1, 6):
if not vacation:
return first_alarm
else:
return second_alarm
if day == 0 or day == 6:
if not vacation:
return second_alarm
else:
return "off"
view rawpuzzle29.py hosted with ❤ by GitHub


Puzzle 30

Given a number n, return True if n is in the range 1..10, inclusive.Unless outside_mode is True, in which case return True if the number is less or equal to 1, or greater or equal to 10.

Solution

def int1to10(n, outside_mode):
if outside_mode:
if n <=1 or n >= 10:
return True
else:
return n in range(1, 11)
return False
view rawpuzzle30.py hosted with ❤ by GitHub



Final thoughts

None of the great professional coders made it to the top without hard work. It is a fact. Without great expectations, consider the puzzles shared through this blog post as a way to keep your brain in shape, and also as a first step in setting a corner stone for your coder journey.

Although there are exactly 30 puzzles shared in here, I am going to add more in the near future. Feel free to contact me, or comment below, if you have any helpful suggestion for the future puzzles.


Source: https://orthodoxpirate.blogspot.com/2021/01/python-puzzles-to-improve-yourself-as.html?ref=idomaster.com


on January 13th, 2021 (11:28 pm)
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