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

1844. Replace All Digits with Characters

Level

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);
    }
}

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