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

def string_times(str, n):
  
    return str*n



##-------------------------------------------------
##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;

def front_times(str, n):
  
    return str[:3]*n



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

def string_bits(str):
  
    return str[::2]



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

def string_splosion(str):
  
    result = ''
    for i in range(len(str)):
        result += str[:i+1]
    return result



##-------------------------------------------------
##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).

def last2(str):

    count = 0
    for i in range(len(str)-2):
        if(str[i:i+2] == str[-2:]):
            count += 1
    return count



##-------------------------------------------------
##Given an array of ints, return the number of 9's in the array.

def array_count9(nums):

    countNine = 0
    for num in nums:
        if(num == 9):
            countNine +=1
    return countNine



##-------------------------------------------------
##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.

def array_front9(nums):
  
    if(9 in nums[:4]):
        return True
    return False



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

def array123(nums):
  
    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



##-------------------------------------------------
##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. 

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

