1844. Replace All Digits with Characters

Problem

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.

Solution (Java)

class Solution {
    public String replaceDigits(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            if (Character.isAlphabetic(c)) {
                sb.append(c);
            } else {
                sb.append((char) (sb.charAt(sb.length() - 1) + Character.getNumericValue(c)));
            }
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).