Python Programming Series # 1 - Warmup Series -1

In this series, we will be learning Python through practice.

All the parts will consist of 10 coding problems for Python Practice. The questions will be in sequence, sorted from low-difficulty level to high-difficulty level.

Series list:-

Warmup-1 : Simple warmup problems to get started, no loops
**Warmup-2 **: Medium warmup string/list problems with loops
String-1 : Basic python string problems -- no loops
List-1 : Basic python list problems -- no loops.
Logic-1 : Basic boolean logic puzzles -- if else and or not
Logic-2 : Medium boolean logic puzzles -- if else and or not
String-2: Medium python string problems -- 1 loop.
List-2 : Medium python list problems -- no loops.

Coding

Warmup-1 > sleep_in

Question:
The parameter weekday is True if it is a weekday, and the parameter vacation is True if we are on vacation. We sleep in if it is not a weekday or we're on vacation. Return True if we sleep in.

Check for these cases:

sleep_in(False, False) → True 
sleep_in(True, False) → False
sleep_in(False, True) → True

Solution:

def sleep_in(weekday, vacation):
    if(not weekday or vacation):
        print("True")
    else:
        print("False")

sleep_in(False, False) #→ True
sleep_in(True, False) #→ False
sleep_in(False, True) #→ True

#----------------------------------------#

Warmup-1 > monkey_trouble

Question:
We have two monkeys, a and b, and the parameters a_smile and b_smile indicate if each is smiling. We are in trouble if they are both smiling or if neither of them is smiling. Return True if we are in trouble.

Check for these cases:

monkey_trouble(True, True) → True
monkey_trouble(False, False) → True
monkey_trouble(True, False) → False

Solution:

def monkey_trouble(a_smile, b_smile):
    if a_smile and b_smile:
        print("True")
    elif not a_smile and not b_smile:
        print("True")
    else:
        print("False")
        
monkey_trouble(True, True) #→ True
monkey_trouble(False, False) #→ True
monkey_trouble(True, False) #→ False

#----------------------------------------#

Warmup-1 > sum_double

Question:
Given two int values, return their sum. Unless the two values are the same, then return double their sum.

Check for these cases:

sum_double(1, 2) → 3
sum_double(3, 2) → 5
sum_double(2, 2) → 8

Solution:

def sum_double(a, b):
    # Store the sum in a local variable
      sum = a + b
      if (a == b):
        sum = sum * 2
      print(sum)
    
#Alternative solution - 
#    if (a == b):
#        print((a+b)*2)
#    else:
#        print(a+b)

sum_double(1, 2) #→ 3
sum_double(3, 2) #→ 5
sum_double(2, 2) #→ 8

#----------------------------------------#

Warmup-1 > diff21

Question:
Given an int n, return the absolute difference between n and 21, except return double the absolute difference if n is over 21.

Check for these cases:

diff21(19) → 2
diff21(10) → 11
diff21(21) → 0

Solution:

def diff21(n):
    diff = 21-n
    if n>21:
        diff = diff*2
    print(diff)

diff21(19) #→ 2
diff21(10) #→ 11
diff21(21) #→ 0

#----------------------------------------#

Warmup-1 > parrot_trouble

Question:
We have a loud talking parrot. The "hour" parameter is the current hour time in the range 0..23. We are in trouble if the parrot is talking and the hour is before 7 or after 20. Return True if we are in trouble.

Check for these cases:

parrot_trouble(True, 6) → True
parrot_trouble(True, 7) → False
parrot_trouble(False, 6) → False

Solution:

def parrot_trouble(talking, hour):
    if(talking and (hour<7 or hour>20)):
        print("True")
    else:
        print("False")
        
parrot_trouble(True, 6) #→ True
parrot_trouble(True, 7) #→ False
parrot_trouble(False, 6) #→ False

#----------------------------------------#

Warmup-1 > makes10

Question:
Given 2 ints, a and b, return True if one if them is 10 or if their sum is 10.

Check for these cases:

makes10(9, 10) → True
makes10(9, 9) → False
makes10(1, 9) → True

Solution:

def makes10(a, b):
    if((a == 10) or (b == 10) or (a+b == 10)):
        print("True")
    else:
        print("False")
        
makes10(9, 10) #→ True
makes10(9, 9) #→ False
makes10(1, 9) #→ True

#----------------------------------------#

Warmup-1 > near_hundred

Question:
Given an int n, return True if it is within 10 of 100 or 200. Note: abs(num) computes the absolute value of a number.

Check for these cases:

near_hundred(93) → True
near_hundred(90) → True
near_hundred(89) → False

Solution:

def near_hundred(n):
    diff1 = abs(100-n)
    diff2 = abs(200-n)
    if(diff1 <= 10 or diff2 <=10):
        print("True")
    else:
        print("False")
        
near_hundred(93) #→ True
near_hundred(90) #→ True
near_hundred(89) #→ False
near_hundred(190) #→ True

#----------------------------------------#

Warmup-1 > pos_neg

Question:
Given 2 int values, return True if one is negative and one is positive. Except if the parameter "negative" is True, then return True only if both are negative.

Check for these cases:

pos_neg(1, -1, False) → True
pos_neg(-1, 1, False) → True
pos_neg(-4, -5, True) → True

Solution:

def pos_neg(a, b, negative):
    if negative: 
        print(a<0 and b<0)
    else:
        print((a<0 and b>0) or (a>0 and b<0))
        
pos_neg(1, -1, False) #→ True
pos_neg(-1, 1, False) #→ True
pos_neg(-4, -5, True) #→ True

#----------------------------------------#

Warmup-1 > not_string

Question:
Given a string, return a new string where "not " has been added to the front. However, if the string already begins with "not", return the string unchanged.

Check for these cases:

not_string('candy') → 'not candy'
not_string('x') → 'not x'
not_string('not bad') → 'not bad'

Solution:

def not_string(str):
    if(len(str) >= 3 and str[0:3] == 'not'):
        print(str)
    else:
        print('not '+str)
        
# str[:3] goes from the start of the string up to but not
# including index 3
    
not_string('candy') #→ 'not candy'
not_string('x') #→ 'not x'
not_string('not bad') #→ 'not bad'

#----------------------------------------#

#----------------------------------------

Warmup-1 > missing_char

Question:
Given a non-empty string and an int n, return a new string where the char at index n has been removed. The value of n will be a valid index of a char in the original string (i.e. n will be in the range 0..len(str)-1 inclusive).

Check for these cases:

missing_char('kitten', 1) → 'ktten'
missing_char('kitten', 0) → 'itten'
missing_char('kitten', 4) → 'kittn'

Solution:

def missing_char(str, n):
    front  = str[:n]  # upto str[n-1] 
    back = str[n+1:]  # from n+1 to end of the string
    print(front + back)
    
missing_char('kitten', 1) #→ 'ktten'
missing_char('kitten', 0) #→ 'itten'
missing_char('kitten', 4) #→ 'kittn'

#----------------------------------------#

Warmup-1 > front_back_reverse

Question:
Given a string, find the reversed string.

Check for these cases:

front_back_rev('code') → 'eodc'
front_back_rev('a') → 'a'
front_back('ab') → 'ba'

Solution:

def front_back_rev(str): 
    rstr = ""  # initialized the unused variable.
    for i in range(len(str)):
        rstr += str[(len(str)-1)-i]
    print(rstr)

front_back_rev('code') #→ 'edoc'
front_back_rev('a') #→ 'a'
front_back_rev('ab') #→ 'ba'

#----------------------------------------#

Warmup-1 > front_back

Question:
Given a string, return a new string where the first and last chars have been exchanged.

Check for these cases:

front_back('code') → 'eodc'
front_back('a') → 'a'
front_back('ab') → 'ba'

Solution:

def front_back(str):
    if(len(str)<=1):  # if the string is of single character, print the same string.
        print(str)
    else:
        front = str[0]  # 1st character
        mid = str[1:len(str)-1]
        last = str[len(str)-1] # last character
        new_str = last + mid + front # string added in reverse order, with middle part same as previous.
        print(new_str)

front_back('code') #→ 'eodc'
front_back('a') #→ 'a'
front_back('ab') #→ 'ba'

#----------------------------------------#

Warmup-1 > front3

Question:
Given a string, we'll say that the front is the first 3 chars of the string. If the string length is less than 3, the front is whatever is there. Return a new string which is 3 copies of the front.

Check for these cases:

front3('Java') → 'JavJavJav'
front3('Chocolate') → 'ChoChoCho'
front3('abc') → 'abcabcabc'

Solution:

def front3(str):
    if(len(str)>=3): # if the same string has 3 or more characters string, print the required string
        print(3 * str[0:3])  # str[0:3] means from str[0] to str[2]
    else:          # if the string has less than 3 character string, print the same string.
        print(3*str)

front3('Java') #→ 'JavJavJav'
front3('Chocolate') #→ 'ChoChoCho'
front3('abc') #→ 'abcabcabc'