Palindromic partitions
Detailed guide and Python implementation for the 'Palindromic partitions' problem.
1. Concept Overview
The 'Palindromic partitions' 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 Palindromic partitions.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Palindromic partitions 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 palindromic_partitions(s) that returns all possible ways to partition the string s such that every substring in the partition is a palindrome. Return a sorted list of partitions, where each partition is a list of strings. Sort by comparing partitions lexicographically.
- •1 <= len(s) <= 16
- •s contains only lowercase English letters
Examples
s = 'aab'
[['a', 'a', 'b'], ['aa', 'b']]
'a','a','b' are all palindromes. 'aa' is a palindrome and 'b' is a palindrome. 'aab' itself is not a palindrome.
s = 'a'
[['a']]
A single character is always a palindrome.
s = 'aba'
[['a', 'b', 'a'], ['aba']]
Both ['a','b','a'] and ['aba'] consist of only palindromic substrings.
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 partition(s):
n = len(s)
dp = [[False] * n for _ in range(n)]
for i in range(n):
for j in range(i + 1):
if s[i] == s[j] and (i - j <= 2 or dp[j + 1][i - 1]):
dp[j][i] = True
res = []
def backtrack(start, path):
if start == n: res.append(list(path)); return
for end in range(start, n):
if dp[start][end]:
path.append(s[start:end+1]); backtrack(end + 1, path); path.pop()
backtrack(0, [])
return resBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def partition(s):
res = []
def dfs(s, path):
if not s: res.append(path); return
for i in range(1, len(s) + 1):
if s[:i] == s[:i][::-1]: dfs(s[i:], path + [s[:i]])
dfs(s, [])
return resAlgorithm Pattern Checklist
When dealing with Recursion data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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