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Abhishek Chaudhary

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Replace All Digits with Characters

You are given a 0-indexed string `s` that has lowercase English letters in its even indices and digits in its odd indices.

There is a function `shift(c, x)`, where `c` is a character and `x` is a digit, that returns the `xth` character after `c`.

• For example, `shift('a', 5) = 'f'` and `shift('x', 0) = 'x'`.

For every odd index `i`, you want to replace the digit `s[i]` with `shift(s[i-1], s[i])`.

Return `s` after replacing all digits. It is guaranteed that `shift(s[i-1], s[i])` will never exceed `'z'`.

Example 1:

Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:

• s[1] -> shift('a',1) = 'b'
• s[3] -> shift('c',1) = 'd'
• s[5] -> shift('e',1) = 'f'

Example 2:

Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:

• s[1] -> shift('a',1) = 'b'
• s[3] -> shift('b',2) = 'd'
• s[5] -> shift('c',3) = 'f'
• s[7] -> shift('d',4) = 'h'

Constraints:

• `1 <= s.length <= 100`
• `s` consists only of lowercase English letters and digits.
• `shift(s[i-1], s[i]) <= 'z'` for all odd indices `i`.

SOLUTION:

``````class Solution:
def replaceDigits(self, s: str) -> str:
return "".join([chr(ord(s[i - 1]) + int(c)) if i & 1 else c for i, c in enumerate(s)])
``````