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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.

I have achieved this with the following code using regular expressions:

import re


def count_code(str):
    exp = '^co[a-z|A-Z]e$'
    count = 0
    for i in range(len(str) - 1):
        if re.match(exp, str[i:i + 4]):
            count = count + 1

    return count

print count_code('aaacodebbb')  # prints 1
print count_code('codexxcode')  # prints 2
print count_code('cozexxcope')  # prints 2

Is there any other way of achieving this without using regular expression

 solution

this is a simple and clean solution for this problem:

  def count_code(str):
      count = 0
      for i in range(len(str)):
        if str[i:i+2] == "co" and str[i+3:i+4] == "e":
          count+=1
      return count

def count_code(str):
  a = 0
  for i in range(len(str) - 3):
    if str[i:i+2] + str[i+3] == 'coe':
      a += 1
  return a

Using Python String Method ‘count’

 def count_code1(str):
       counts=0
       for i in range(97,123):   #all the lowercase ASCII characters
        count+= str.count('co'+chr(i)+'e')
       return counts 

This should work too:

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

Answer

One way is you can make every possible string with co*e where * is any alphabet

Like

x=["co"+i+"e" for i in string.lowercase]

Then iterate

for i in x:
    if i in <your string>:
        count+=<your string>.count(i)

You can try this:

def count_code(str):
    x=["co"+i+"e" for i in str.lower()]
    count = 0
    index = 0
    for i in x:
        if i in str[index:]:
            index = str.find(i)+1
            count+=1
    return count

print count_code('aaacodebbb')  # prints 1
print count_code('codexxcode')  # prints 2
print count_code('cozexxcope')  # prints 2    

You can try also : using Python String Method ‘count’

 def count_code1(str):
       counts=0
       for i in range(97,123):   #all the lowercase ASCII characters
        count+= str.count('co'+chr(i)+'e')
       return counts