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

# 1844. Replace All Digits with Characters

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 -> shift(‘a’,1) = ‘b’
• s -> shift(‘c’,1) = ‘d’
• s -> shift(‘e’,1) = ‘f’

Example 2:

Input: s = “a1b2c3d4e”

Output: “abbdcfdhe”

Explanation: The digits are replaced as follows:

• s -> shift(‘a’,1) = ‘b’
• s -> shift(‘b’,2) = ‘d’
• s -> shift(‘c’,3) = ‘f’
• s -> 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);
}
}