LeetCode 1844. Replace All Digits with Characters

Description

https://leetcode.com/problems/replace-all-digits-with-characters/

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.

Explanation

Iterate the string, if at odd position, keep the original character, otherwise at even position, base on the value to replace with a new character.

Python Solution

class Solution:
    def replaceDigits(self, s: str) -> str:
        result = ""
        
        for i, c in enumerate(s):
            if i % 2 == 0:
                result += c
            else:                
                result += chr(ord(s[i - 1]) + int(c)) 
        
        return result
  • Time Complexity: O(N).
  • Space Complexity: O(N).

Leave a Reply

Your email address will not be published.