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Marc Peejay V. Viernes

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# 📝 Exploring Palindromic Partitioning: Solving the "Palindrome Partitioning" Problem 📝

Welcome to another journey through the realm of algorithms and problem-solving! Today, we're delving into the intriguing "Palindrome Partitioning" problem, a medium-level 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:

1. Example 1:

• Input: `s = "aab"`
• Output: `[["a","a","b"],["aa","b"]]`
2. Example 2:

• Input: `s = "a"`
• Output: `[["a"]]`

### Constraints

• 1 <= s.length <= 16
• `s` contains only lowercase English letters.

## Solution Approach

To tackle this problem, we utilize backtracking through depth-first 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 lists `results` and `part` to store the final partition results and the current partition being explored, respectively.
• We define a `depth_first_search` function that implements depth-first search (DFS) for backtracking. It recursively explores all possible partitionings starting from a given index `i`.
• Within the `depth_first_search` function, we iterate over possible substring lengths starting from index `i` 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 string `s` from index `l` to `r` (inclusive) is a palindrome.

### Function Invocation

• The `depth_first_search` function is initially called with the starting index `0` to initiate the backtracking process.
• The final list of palindrome partitionings stored in `results` is returned by the `partition` function.

## Complexity Analysis

• Time Complexity: O(N * 2^N), where N is the length of string `s`. This is the worst-case 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 depth-first 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! 🚀