Leetcode #1576: Replace All ?'s to Avoid Consecutive Repeating Characters
In this guide, we solve Leetcode #1576 Replace All ?'s to Avoid Consecutive Repeating 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
Given a string s containing only lowercase English letters and the '?' character, convert all the '?' characters into lowercase letters such that the final string does not contain any consecutive repeating characters. You cannot modify the non '?' characters.
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 = "?zs"
Output: "azs"
Explanation: There are 25 solutions for this problem. From "azs" to "yzs", all are valid. Only "z" is an invalid modification as the string will consist of consecutive repeating characters in "zzs".
Python Solution
class Solution:
def modifyString(self, s: str) -> str:
s = list(s)
n = len(s)
for i in range(n):
if s[i] == "?":
for c in "abc":
if (i and s[i - 1] == c) or (i + 1 < n and s[i + 1] == c):
continue
s[i] = c
break
return "".join(s)
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
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.