Replace All Digits with Characters Using Python

Suppose we have an alphanumeric string s that contains lowercase English letters in its even positions and digits in its odd positions. Consider an operation shift(c, x), where c is any character and x is a number (digit), this will find the xth character after c. So, for example, shift('p', 5) = 'u' and shift('a', 0) = 'a'. Now for every odd index i, we want to replace the digit s[i] with shift(s[i-1], s[i]). We have to find s after replacing all digits.

So, if the input is like s = "a2b1d4f3h2", then the output will be "acbcdhfihj" because

• shift('a', 2) = 'c'

• shift('b', 1) = 'c'

• shift('d', 4) = 'h'

• shift('f', 3) = 'i'

• shift('h', 2) = 'j'

To solve this, we will follow these steps −

• res:= blank string

• for i in range 0 to size of s, do

  • if s[i] is a digit, then

    • res := res concatenate character from (ASCII s[i] + ASCII of s[i-1])

  • otherwise,

    • res := res concatenate s[i]

• return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

def solve(s):
   res=""
   for i in range(len(s)):
      if s[i].isdigit():
         res+= chr(int(s[i])+ord(s[i-1]))
      else:
         res+=s[i]
   return res
s = "a2b1d4f3h2"
print(solve(s))

Input

"a2b1d4f3h2"

Output

acbcdhfihj