README_EN.md

1844. Replace All Digits with Characters

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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

Python3

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)

Java

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

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