warmup_2.py

#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Wed Mar 13 13:57:26 2019

@author: YaoJunyan
"""

'''
string_times

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):
  return str*n

'''
front_times

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('Abc', 3) → 'AbcAbcAbc'
'''
def front_times(str, n):
  if len(str)>=3:
    return str[0:3]*n
  else:
    return str*n

'''
string_bits

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

'''
string_splosion

Given a non-empty string like "Code" return a string like "CCoCodCode".


string_splosion('Code') → 'CCoCodCode'
string_splosion('abc') → 'aababc'
string_splosion('ab') → 'aab'
'''
def string_splosion(str):
  result =''
  for i in range(len(str)+1):
    result =result + str[0:i]
  return result

'''last2

Given a string, return the count of the number of times that a substring length 2 appears in the string and also as the last 2 chars of the string, so "hixxxhi" yields 1 (we won't count the end substring).


last2('hixxhi') → 1
last2('xaxxaxaxx') → 1
last2('axxxaaxx') → 2
'''
def last2(str):
  count =0
  pattern= str[-2:]
  flag=True
  start=0
  while flag:
    a = str.find(pattern,start)  # find() returns -1 if the word is not found, 
    if a==-1:
      flag=False
    else:
      count+=1        
      start=a+1
  return count-1

'''array_count9

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):
  count=0
  for i in range(len(nums)):
    if nums[i]==9:
      count += 1
  return count

'''
array_front9

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):
  if len(nums)<=4:
    return 9 in nums
  elif 9 in nums[0:4]:
    return True
  else:
    return False

'''
array123

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):
  numstr= ''.join([str(i) for i in nums])
  return '123' in numstr

'''
string_match

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):
  count=0
  shorter = min(len(a),len(b))
  for i in range(shorter-1):
    if a[i:i+2]==b[i:i+2]:
      count +=1
    else:
      count = count+0
  return count