String-2.py

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Wed Mar 13 14:43:10 2019

@author: YaoJunyan
"""

'''double_char

Given a string, return a string where for every char in the original, there are two chars.


double_char('The') → 'TThhee'
double_char('AAbb') → 'AAAAbbbb'
double_char('Hi-There') → 'HHii--TThheerree'
'''
def double_char(str):
  strlist= list(str)
  double= [i*2 for i in strlist]
  return ''.join(double)

'''count_hi


Return the number of times that the string "hi" appears anywhere in the given string.


count_hi('abc hi ho') → 1
count_hi('ABChi hi') → 2
count_hi('hihi') → 2

'''
def count_hi(str):
  count=0
  for i in range(len(str)):
    if str[i:i+2]=='hi':
      count  +=1
  return count

'''cat_dog


Return True if the string "cat" and "dog" appear the same number of times in the given string.


cat_dog('catdog') → True
cat_dog('catcat') → False
cat_dog('1cat1cadodog') → True

'''
def cat_dog(str):
  countcat=0
  countdog =0
  for i in range(len(str)-2):
    if str[i:i+3]=='cat':
      countcat +=1
    if str[i:i+3] =='dog':
      countdog +=1
  return countcat== countdog
    
'''count_code

Return the number of times that the string "code" appears anywhere in the given string, except we'll accept any letter for the 'd', so "cope" and "cooe" count.


count_code('aaacodebbb') → 1
count_code('codexxcode') → 2
count_code('cozexxcope') → 2

'''
def count_code(str):
  count =0
  for i in range(len(str)-3):
    if str[i:i+2] =='co' and str[i+3]=='e':
      count +=1
  return count


'''
end_other

Given two strings, return True if either of the strings appears at the very end of the other string, ignoring upper/lower case differences (in other words, the computation should not be "case sensitive"). Note: s.lower() returns the lowercase version of a string.


end_other('Hiabc', 'abc') → True
end_other('AbC', 'HiaBc') → True
end_other('abc', 'abXabc') → True

'''

def end_other(a, b):
  if len(a)>len(b):
    if a[-len(b):].lower()==b.lower():
      return True
    else:
      return False
  else:
    if b[-len(a):].lower()==a.lower():
      return True
    else:
      return False


'''xyz_three

Return True if the given string contains an appearance of "xyz" where the xyz is not directly preceeded by a period (.). So "xxyz" counts but "x.xyz" does not.


xyz_there('abcxyz') → True
xyz_there('abc.xyz') → False
xyz_there('xyz.abc') → True
'''
def xyz_there(str):
  #this is the anser i found from stackoverflow
  return str.count('xyz')- str.count('.xyz')>0