Welcome to another journey through the realm of algorithms and problemsolving! Today, we're delving into the intriguing "Palindrome Partitioning" problem, a mediumlevel challenge that tests our ability to partition a string into substrings, each of which forms a palindrome.
Problem Overview
Given a string s
, the task is to partition it such that every substring of the partition is a palindrome. Our goal is to return all possible palindrome partitionings of s
.
Examples
Let's explore a few examples to better understand the problem:

Example 1:

Input:
s = "aab"

Output:
[["a","a","b"],["aa","b"]]

Input:

Example 2:

Input:
s = "a"

Output:
[["a"]]

Input:
Constraints
 1 <= s.length <= 16

s
contains only lowercase English letters.
Solution Approach
To tackle this problem, we utilize backtracking through depthfirst search (DFS) to explore all possible partitionings of the input string. Let's dive into the solution code and understand it step by step.
class Solution:
def partition(self, s: str) > List[List[str]]:
results, part = [], []
def depth_first_search(i):
if i >= len(s):
results.append(part.copy())
return
for j in range(i, len(s)):
if self.isPalindrome(s, i, j):
part.append(s[i : j + 1])
depth_first_search(j + 1)
part.pop()
depth_first_search(0)
return results
def isPalindrome(self, s, l, r):
while l < r:
if s[l] != s[r]:
return False
l, r = l + 1, r  1
return True
Solution Explanation
Overview
 We use a backtracking approach to explore all possible palindrome partitionings of the input string
s
.  The
partition
function initializes empty listsresults
andpart
to store the final partition results and the current partition being explored, respectively.  We define a
depth_first_search
function that implements depthfirst search (DFS) for backtracking. It recursively explores all possible partitionings starting from a given indexi
.  Within the
depth_first_search
function, we iterate over possible substring lengths starting from indexi
up to the end of the string.  For each substring, we check if it is a palindrome using the
isPalindrome
helper function.  If a palindrome substring is found, it is added to the current partition (
part
), and further exploration continues recursively from the next index (j + 1
).  If the end of the string is reached (
i >= len(s)
), the current partition is a valid palindrome partition, and a copy of it is added to the final results (results
).  After exploring all possibilities, the function returns the final list of palindrome partitionings.
Helper Function
 The
isPalindrome
helper function checks if a substring of the input strings
from indexl
tor
(inclusive) is a palindrome.
Function Invocation
 The
depth_first_search
function is initially called with the starting index0
to initiate the backtracking process.  The final list of palindrome partitionings stored in
results
is returned by thepartition
function.
Complexity Analysis

Time Complexity: O(N * 2^N), where N is the length of string
s
. This is the worstcase time complexity when all possible substrings are palindromes. 
Space Complexity: O(N), where N is the length of the string
s
. This space will be used to store the recursion stack.
Conclusion
The "Palindrome Partitioning" problem challenges us to efficiently partition a string into substrings, ensuring that each substring is a palindrome. By leveraging backtracking through depthfirst search, we explore all possible partitionings and validate palindromic substrings along the way. While dynamic programming offers a potential optimization for future enhancements, the current approach provides a clear and concise solution to the problem. This problem showcases the power of algorithmic techniques in solving complex string manipulation challenges.
For further exploration and practice, visit the "Palindrome Partitioning" problem on LeetCode. Happy coding! 🚀
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