Remove adjacent duplicates recursively
Detailed guide and Python implementation for the 'Remove adjacent duplicates recursively' problem.
1. Concept Overview
The 'Remove adjacent duplicates recursively' problem is a key challenge in the Recursion section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Remove adjacent duplicates recursively.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Remove adjacent duplicates recursively carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function remove_adjacent_duplicates(s) that recursively removes all adjacent duplicate characters from the string s until no adjacent duplicates remain. In each pass, remove all pairs of consecutive identical characters, then repeat the process on the resulting string until it stabilizes. Return the final string.
- •0 <= len(s) <= 1000
- •s contains only lowercase English letters
Examples
s = 'aabccba'
'a'
First pass: remove 'aa' and 'cc' -> 'bba'. Second pass: remove 'bb' -> 'a'. No more adjacent duplicates.
s = 'abcddcba'
''
Remove 'dd' -> 'abccba'. Remove 'cc' -> 'abba'. Remove 'bb' -> 'aa'. Remove 'aa' -> ''. Empty string.
s = 'abcd'
'abcd'
No adjacent duplicates exist, so the string remains unchanged.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def remove_adj(s):
def solve(string):
if not string: return ""
res, i, n = [], 0, len(string)
while i < n:
if (i < n - 1 and string[i] != string[i+1]) and (i == 0 or string[i] != string[i-1]): res.append(string[i])
elif i == n - 1 and (i == 0 or string[i] != string[i-1]): res.append(string[i])
i += 1
new_s = "".join(res)
return solve(new_s) if len(new_s) != len(string) else new_s
return solve(s)Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def remove_duplicates(s):
if not s or len(s) == 1: return s
if s[0] == s[1]:
i = 0
while i < len(s) - 1 and s[i] == s[i+1]: i += 1
return remove_duplicates(s[i+1:])
rem = remove_duplicates(s[1:])
if rem and rem[0] == s[0]: return remove_duplicates(rem[1:])
return s[0] + remAlgorithm Pattern Checklist
When dealing with Recursion data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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