Formatted question description: https://leetcode.ca/all/1844.html

# 1844. Replace All Digits with Characters

## Level

Easy

## Description

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 `x-th`

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

Loop over odd indices of `s`

and convert the characters to the shifted characters. Then return the modified string.

```
class Solution {
public String replaceDigits(String s) {
char[] array = s.toCharArray();
int length = array.length;
for (int i = 1; i < length; i += 2) {
char letter = array[i - 1];
int digit = array[i] - '0';
array[i] = (char) (letter + digit);
}
return new String(array);
}
}
```