Ex1

Given a string and a non-negative int n, return a larger string that is n copies of the original string.

string_times(‘Hi’, 2) → ‘HiHi’
string_times(‘Hi’, 3) → ‘HiHiHi’
string_times(‘Hi’, 1) → ‘Hi’

def string_times(str, n):

Answer

def string_times(str, n):
  result = ""
  for i in range(n): 
    result = result + str  # could use += here
  return result

Tests

string_times('Hi', 2)

## 'HiHi'

string_times('Hi', 3)

## 'HiHiHi'

string_times('Hi', 1)

## 'Hi'

Ex2

Given a string and a non-negative int n, we’ll say that the front of the string is the first 3 chars, or whatever is there if the string is less than length 3. Return n copies of the front;

front_times(‘Chocolate’, 2) → ‘ChoCho’
front_times(‘Chocolate’, 3) → ‘ChoChoCho’
front_times(‘No’,2) → ’NoNo"

def front_times(str, n):

Answer

def front_times(str, n):
  front_len = 3
  if front_len > len(str):
    front_len = len(str)
  front = str[:front_len]
  
  result = ""
  for i in range(n):
    result = result + front
  return result

Tests

front_times('Chocolate', 2)

## 'ChoCho'

front_times('Chocolate', 3)

## 'ChoChoCho'

front_times('No',2)

## 'NoNo'

Ex3

Given a string, return a new string made of every other char starting with the first, so “Hello” yields “Hlo”.

string_bits(‘Hello’) → ‘Hlo’
string_bits(‘Hi’) → ‘H’
string_bits(‘Heeololeo’) → ‘Hello’

def string_bits(str):

Answer

def string_bits(str):
  result = ""

  for i in range(len(str)):
    if i % 2 == 0:
      result = result + str[i]
  return result

Tests

string_bits('Hello')

## 'Hlo'

string_bits('Hi')

## 'H'

string_bits('Heeololeo')

## 'Hello'

Ex4

Given an array of ints, return the number of 9’s in the array.

array_count9([1, 2, 9]) → 1
array_count9([1, 9, 9]) → 2
array_count9([1, 9, 9, 3, 9]) → 3

def array_count9(nums):

Answer

def array_count9(nums):
  count = 0
  # Standard loop to look at each value
  for num in nums:
    if num == 9:
      count = count + 1
      
  return count    

Tests

array_count9([1, 2, 9])

## 1

array_count9([1, 9, 9])

## 2

array_count9([1, 9, 9, 3, 9])

## 3

Ex5

Given an array of ints, return True if one of the first 4 elements in the array is a 9. The array length may be less than 4.

array_front9([1, 2, 9, 3, 4]) → True
array_front9([1, 2, 3, 4, 9]) → False
array_front9([1, 2, 3, 4, 5]) → False

def array_front9(nums):

Answer

def array_front9(nums):
  # First figure the end for the loop
  end = len(nums)
  if end > 4:
    end = 4
  
  for i in range(end):  # loop over index [0, 1, 2, 3]
    if nums[i] == 9:
      return True
  return False

Tests

array_front9([1, 2, 9, 3, 4])

## True

array_front9([1, 2, 3, 4, 9])

## False

array_front9([1,9])

## True

Ex6

Given an array of ints, return True if the sequence of numbers 1, 2, 3 appears in the array somewhere.

array123([1, 1, 2, 3, 1]) → True
array123([1, 1, 2, 4, 1]) → False
array123([1, 1, 2, 1, 2, 3]) → True

def array123(nums):

Answer

def array123(nums):
  # Note: iterate with length-2, so can use i+1 and i+2 in the loop
  for i in range(len(nums)-2):
    if nums[i]==1 and nums[i+1]==2 and nums[i+2]==3:
      return True
  return False

Tests

array123([1, 1, 2, 3, 1])

## True

array123([1, 1, 2, 4, 1])

## False

array123([1, 1, 2, 1, 2, 3])

## True

Ex7

Given 2 strings, a and b, return the number of the positions where they contain the same length 2 substring. So “xxcaazz” and “xxbaaz” yields 3, since the “xx”, “aa”, and “az” substrings appear in the same place in both strings.

string_match(‘xxcaazz’, ‘xxbaaz’) → 3
string_match(‘abc’, ‘abc’) → 2
string_match(‘abc’, ‘axc’) → 0

def string_match(a, b):

Answer

def string_match(a, b):
  # Figure which string is shorter.
  shorter = min(len(a), len(b))
  count = 0
  
  # Loop i over every substring starting spot.
  # Use length-1 here, so can use char str[i+1] in the loop
  for i in range(shorter-1):
    a_sub = a[i:i+2]
    b_sub = b[i:i+2]
    if a_sub == b_sub:
      count = count + 1

  return count

Tests

string_match('xxcaazz', 'xxbaaz')

## 3

string_match('abc', 'abc')

## 2

string_match('abc', 'axc')

## 0

Ex8

Given a string s, return the number of vowels in the string. Exclude ‘y’, but include both upper and lower case vowels.

vowel_count(‘Help, I am drowning’) -> 5 vowel_count(‘abCdEf’) -> 2

def vowel_count(s):

Answer

def vowel_count(s):
  vowels = "aeiouAEIOU"
  count = 0
  for i in s:
    if i in vowels:
      count = count + 1
  return count    

Tests

vowel_count('abcdef')

## 2

vowel_count('Abcdef')

## 2