Leetcode #1957: Delete Characters to Make Fancy String
In this guide, we solve Leetcode #1957 Delete Characters to Make Fancy String 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
A fancy string is a string where no three consecutive characters are equal. Given a string s, delete the minimum possible number of characters from s to make it fancy.
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 = "leeetcode"
Output: "leetcode"
Explanation:
Remove an 'e' from the first group of 'e's to create "leetcode".
No three consecutive characters are equal, so return "leetcode".
Python Solution
class Solution:
def makeFancyString(self, s: str) -> str:
ans = []
for i, c in enumerate(s):
if i < 2 or c != s[i - 1] or c != s[i - 2]:
ans.append(c)
return "".join(ans)
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.