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Function Practice Exercises - Solutions¶
Problems are arranged in increasing difficulty:
- Warmup - these can be solved using basic comparisons and methods
- Level 1 - these may involve if/then conditional statements and simple methods
- Level 2 - these may require iterating over sequences, usually with some kind of loop
- Challenging - these will take some creativity to solve
WARMUP SECTION:¶
LESSER OF TWO EVENS: Write a function that returns the lesser of two given numbers if both numbers are even, but returns the greater if one or both numbers are odd¶
lesser_of_two_evens(2,4) --> 2
lesser_of_two_evens(2,5) --> 5
In [1]:
def lesser_of_two_evens(a,b):
if a%2 == 0 and b%2 == 0:
return min(a,b)
else:
return max(a,b)
In [2]:
# Check
lesser_of_two_evens(2,4)
Out[2]:
2
In [3]:
# Check
lesser_of_two_evens(2,5)
Out[3]:
5
ANIMAL CRACKERS: Write a function takes a two-word string and returns True if both words begin with same letter¶
animal_crackers('Levelheaded Llama') --> True
animal_crackers('Crazy Kangaroo') --> False
In [4]:
def animal_crackers(text):
wordlist = text.split()
return wordlist[0][0] == wordlist[1][0]
In [5]:
# Check
animal_crackers('Levelheaded Llama')
Out[5]:
True
In [6]:
# Check
animal_crackers('Crazy Kangaroo')
Out[6]:
False
MAKES TWENTY: Given two integers, return True if the sum of the integers is 20 or if one of the integers is 20. If not, return False¶
makes_twenty(20,10) --> True
makes_twenty(12,8) --> True
makes_twenty(2,3) --> False
In [7]:
def makes_twenty(n1,n2):
return (n1+n2)==20 or n1==20 or n2==20
In [8]:
# Check
makes_twenty(20,10)
Out[8]:
True
In [9]:
# Check
makes_twenty(12,8)
Out[9]:
True
In [10]:
#Check
makes_twenty(2,3)
Out[10]:
False
LEVEL 1 PROBLEMS¶
OLD MACDONALD: Write a function that capitalizes the first and fourth letters of a name¶
old_macdonald('macdonald') --> MacDonald
Note: 'macdonald'.capitalize() returns 'Macdonald'
In [11]:
def old_macdonald(name):
if len(name) > 3:
return name[:3].capitalize() + name[3:].capitalize()
else:
return 'Name is too short!'
In [12]:
# Check
old_macdonald('macdonald')
Out[12]:
'MacDonald'
MASTER YODA: Given a sentence, return a sentence with the words reversed¶
master_yoda('I am home') --> 'home am I'
master_yoda('We are ready') --> 'ready are We'
In [13]:
def master_yoda(text):
return ' '.join(text.split()[::-1])
In [14]:
# Check
master_yoda('I am home')
Out[14]:
'home am I'
In [15]:
# Check
master_yoda('We are ready')
Out[15]:
'ready are We'
ALMOST THERE: Given an integer n, return True if n is within 10 of either 100 or 200¶
almost_there(90) --> True
almost_there(104) --> True
almost_there(150) --> False
almost_there(209) --> True
NOTE: abs(num) returns the absolute value of a number
In [16]:
def almost_there(n):
return ((abs(100 - n) <= 10) or (abs(200 - n) <= 10))
In [17]:
# Check
almost_there(90)
Out[17]:
True
In [18]:
# Check
almost_there(104)
Out[18]:
True
In [19]:
# Check
almost_there(150)
Out[19]:
False
In [20]:
# Check
almost_there(209)
Out[20]:
True
LEVEL 2 PROBLEMS¶
FIND 33:¶
Given a list of ints, return True if the array contains a 3 next to a 3 somewhere.
has_33([1, 3, 3]) → True
has_33([1, 3, 1, 3]) → False
has_33([3, 1, 3]) → False
In [21]:
def has_33(nums):
for i in range(0, len(nums)-1):
# nicer looking alternative in commented code
#if nums[i] == 3 and nums[i+1] == 3:
if nums[i:i+2] == [3,3]:
return True
return False
In [22]:
# Check
has_33([1, 3, 3])
Out[22]:
True
In [23]:
# Check
has_33([1, 3, 1, 3])
Out[23]:
False
In [24]:
# Check
has_33([3, 1, 3])
Out[24]:
False
PAPER DOLL: Given a string, return a string where for every character in the original there are three characters¶
paper_doll('Hello') --> 'HHHeeellllllooo'
paper_doll('Mississippi') --> 'MMMiiissssssiiippppppiii'
In [25]:
def paper_doll(text):
result = ''
for char in text:
result += char * 3
return result
In [26]:
# Check
paper_doll('Hello')
Out[26]:
'HHHeeellllllooo'
In [27]:
# Check
paper_doll('Mississippi')
Out[27]:
'MMMiiissssssiiissssssiiippppppiii'
BLACKJACK: Given three integers between 1 and 11, if their sum is less than or equal to 21, return their sum. If their sum exceeds 21 and there's an eleven, reduce the total sum by 10. Finally, if the sum (even after adjustment) exceeds 21, return 'BUST'¶
blackjack(5,6,7) --> 18
blackjack(9,9,9) --> 'BUST'
blackjack(9,9,11) --> 19
In [28]:
def blackjack(a,b,c):
if sum((a,b,c)) <= 21:
return sum((a,b,c))
elif sum((a,b,c)) <=31 and 11 in (a,b,c):
return sum((a,b,c)) - 10
else:
return 'BUST'
In [29]:
# Check
blackjack(5,6,7)
Out[29]:
18
In [30]:
# Check
blackjack(9,9,9)
Out[30]:
'BUST'
In [31]:
# Check
blackjack(9,9,11)
Out[31]:
19
SUMMER OF '69: Return the sum of the numbers in the array, except ignore sections of numbers starting with a 6 and extending to the next 9 (every 6 will be followed by at least one 9). Return 0 for no numbers.¶
summer_69([1, 3, 5]) --> 9
summer_69([4, 5, 6, 7, 8, 9]) --> 9
summer_69([2, 1, 6, 9, 11]) --> 14
In [32]:
def summer_69(arr):
total = 0
add = True
for num in arr:
while add:
if num != 6:
total += num
break
else:
add = False
while not add:
if num != 9:
break
else:
add = True
break
return total
In [33]:
# Check
summer_69([1, 3, 5])
Out[33]:
9
In [34]:
# Check
summer_69([4, 5, 6, 7, 8, 9])
Out[34]:
9
In [35]:
# Check
summer_69([2, 1, 6, 9, 11])
Out[35]:
14
CHALLENGING PROBLEMS¶
SPY GAME: Write a function that takes in a list of integers and returns True if it contains 007 in order¶
spy_game([1,2,4,0,0,7,5]) --> True
spy_game([1,0,2,4,0,5,7]) --> True
spy_game([1,7,2,0,4,5,0]) --> False
In [36]:
def spy_game(nums):
code = [0,0,7,'x']
for num in nums:
if num == code[0]:
code.pop(0) # code.remove(num) also works
return len(code) == 1
In [37]:
# Check
spy_game([1,2,4,0,0,7,5])
Out[37]:
True
In [38]:
# Check
spy_game([1,0,2,4,0,5,7])
Out[38]:
True
In [39]:
# Check
spy_game([1,7,2,0,4,5,0])
Out[39]:
False
COUNT PRIMES: Write a function that returns the number of prime numbers that exist up to and including a given number¶
count_primes(100) --> 25
By convention, 0 and 1 are not prime.
In [40]:
def count_primes(num):
primes = [2]
x = 3
if num < 2: # for the case of num = 0 or 1
return 0
while x <= num:
for y in range(3,x,2): # test all odd factors up to x-1
if x%y == 0:
x += 2
break
else:
primes.append(x)
x += 2
print(primes)
return len(primes)
In [41]:
# Check
count_primes(100)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
Out[41]:
25
BONUS: Here's a faster version that makes use of the prime numbers we're collecting as we go!
In [42]:
def count_primes2(num):
primes = [2]
x = 3
if num < 2:
return 0
while x <= num:
for y in primes: # use the primes list!
if x%y == 0:
x += 2
break
else:
primes.append(x)
x += 2
print(primes)
return len(primes)
In [43]:
count_primes2(100)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
Out[43]:
25
Just for fun, not a real problem :)¶
PRINT BIG: Write a function that takes in a single letter, and returns a 5x5 representation of that letter¶
print_big('a')
out: *
* *
*****
* *
* *
HINT: Consider making a dictionary of possible patterns, and mapping the alphabet to specific 5-line combinations of patterns.
For purposes of this exercise, it's ok if your dictionary stops at "E".
In [44]:
def print_big(letter):
patterns = {1:' * ',2:' * * ',3:'* *',4:'*****',5:'**** ',6:' * ',7:' * ',8:'* * ',9:'* '}
alphabet = {'A':[1,2,4,3,3],'B':[5,3,5,3,5],'C':[4,9,9,9,4],'D':[5,3,3,3,5],'E':[4,9,4,9,4]}
for pattern in alphabet[letter.upper()]:
print(patterns[pattern])
In [45]:
print_big('a')
* * * ***** * * * *