##### Welcome to Subscribe On Youtube

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[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);
}
}

############

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);
}
}

• // OJ: https://leetcode.com/problems/replace-all-digits-with-characters/
// Time: O(N)
// Space: O(1)
class Solution {
public:
string replaceDigits(string s) {
for (int i = 1; i < s.size(); 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)

############

# 1844. Replace All Digits with Characters
# https://leetcode.com/problems/replace-all-digits-with-characters/

class Solution:
def replaceDigits(self, s: str) -> str:
res = []

for i, c in enumerate(s):
if c.isnumeric():
res.append(chr(ord(s[i-1]) + int(c)))
else:
res.append(c)

return "".join(res)


• 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()
}
}