Leetcode #1844: Replace All Digits with Characters
In this guide, we solve Leetcode #1844 Replace All Digits with Characters in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices. You must perform an operation shift(c, x), where c is a character and x is a digit, that returns the xth character after c.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: String
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
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'
Python Solution
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)
Complexity
The time complexity is , where is the length of the string . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.