# LeetCode 1844. Replace All Digits with Characters

## Description

https://leetcode.com/problems/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`.

## Explanation

Iterate the string, if at odd position, keep the original character, otherwise at even position, base on the value to replace with a new character.

## Python Solution

``````class Solution:
def replaceDigits(self, s: str) -> str:
result = ""

for i, c in enumerate(s):
if i % 2 == 0:
result += c
else:
result += chr(ord(s[i - 1]) + int(c))

return result``````
• Time Complexity: O(N).
• Space Complexity: O(N).