# 1844. Replace All Digits with Characters

## 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 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.

## Solutions

Solution 1: Simulation

Traverse the string, for characters at odd indices, replace them with the character that is a certain number of positions after the previous character.

Finally, return the replaced string.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

• class Solution {
public String replaceDigits(String s) {
char[] cs = s.toCharArray();
for (int i = 1; i < cs.length; i += 2) {
cs[i] = (char) (cs[i - 1] + (cs[i] - '0'));
}
return String.valueOf(cs);
}
}

• class Solution {
public:
string replaceDigits(string s) {
int n = s.size();
for (int i = 1; i < n; i += 2) {
s[i] = s[i - 1] + s[i] - '0';
}
return s;
}
};

• class Solution:
def replaceDigits(self, s: str) -> str:
s = list(s)
for i in range(1, len(s), 2):
s[i] = chr(ord(s[i - 1]) + int(s[i]))
return ''.join(s)


• func replaceDigits(s string) string {
cs := []byte(s)
for i := 1; i < len(s); i += 2 {
cs[i] = cs[i-1] + cs[i] - '0'
}
return string(cs)
}

• function replaceDigits(s: string): string {
const n = s.length;
const ans = [...s];
for (let i = 1; i < n; i += 2) {
ans[i] = String.fromCharCode(ans[i - 1].charCodeAt(0) + Number(ans[i]));
}
return ans.join('');
}


• impl Solution {
pub fn replace_digits(s: String) -> String {
let n = s.len();
let mut ans = s.into_bytes();
let mut i = 1;
while i < n {
ans[i] = ans[i - 1] + (ans[i] - b'0');
i += 2;
}
ans.into_iter().map(char::from).collect()
}
}