131. Palindrome Partitioning - Detailed Explanation

Problem Statement

Description:
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

A palindrome is a string that reads the same forward and backward.

Constraints:

The length of string s is between 1 and 16.
s consists of lowercase English letters only.

Example 1:

Input: s = "aab"
Output: [["a","a","b"], ["aa","b"]]
Explanation:
- One valid partition is splitting into "a", "a", and "b", where each substring is a palindrome.
- Another valid partition is splitting into "aa" and "b", where "aa" is a palindrome.

Example 2:

Input: s = "a"
Output: [["a"]]
Explanation:
- The only possible partition is the string itself.

Example 3:

Input: s = "racecar"
Output: [["r", "a", "c", "e", "c", "a", "r"], ["r", "aceca", "r"], ["racecar"]]
Explanation:
- There are multiple ways to partition "racecar" into palindromic substrings.

Hints Before the Solution

Hint 1: Think about how you can check if a substring is a palindrome in an efficient manner.
Hint 2: Consider using a backtracking approach to explore every possible partition and only keep the partitions where every substring is a palindrome.

Approaches

Approach 1: Brute Force (Generate All Partitions)

Explanation:

Idea:
Generate all possible ways to partition the string into substrings, then check each partition to determine if every substring is a palindrome.
How It Works:
- Enumerate every possible partition (there are $(2^{(n-1)})$ ways to partition a string of length (n)).
- For each partition, verify if every substring is a palindrome.
Drawbacks:
- This approach is conceptually simple but inefficient as it involves generating many invalid partitions.
Complexity:
- Time: Exponential, $O(2^n * n)$ in the worst-case.
- Space: O(n) for recursion and partition storage.

Approach 2: Backtracking (Optimal Approach)

Explanation:

Idea:
Use backtracking to explore possible partitions. At each step, choose a prefix that is a palindrome and recursively partition the remainder of the string.
How It Works:
1. Start with an empty list for the current partition.
2. Iterate through the string and for each index, check if the substring from the current start to that index is a palindrome.
3. If it is, add it to the current partition and recursively process the remaining substring.
4. When the entire string is partitioned, add the current partition to the result list.
Why It Works:
This method only explores valid partitions by making sure each chosen substring is a palindrome before proceeding.
Complexity:
- Time: Worst-case time is exponential $O(2^n)$ since there can be a huge number of partitions.
- Space: O(n) for recursion depth and temporary storage.

Python Code

Below is the Python implementation using a backtracking approach:

Python3

. . . .

Java Code

Below is the Java implementation using the backtracking approach.

Java

. . . .

Complexity Analysis

Time Complexity:
- Worst-Case: $O(2^n * n)$ , because in the worst-case scenario (e.g., when every substring is a palindrome, like in a string of identical characters), we explore nearly all possible partitionings.
Space Complexity:
- O(n) for the recursion stack, plus additional space for storing the current partition, and the final results, which in the worst-case may also be exponential in number.

Step-by-Step Walkthrough & Visual Example

Consider the example s = "aab":

Initial Call:
Start with an empty partition and start = 0.
First Level (start = 0):
- Check substring "a" (indices 0 to 0): It is a palindrome.
  - Add "a" to the current partition → current partition becomes ["a"].
  - Recurse with start = 1.
- Also, check substring "aa" (indices 0 to 1): It is a palindrome.
  - Add "aa" to the current partition → current partition becomes ["aa"].
  - Recurse with start = 2.
- Substring "aab" (indices 0 to 2) is not a palindrome.
Second Level (for partition ["a"], start = 1):
- Check substring "a" (indices 1 to 1): It is a palindrome.
  - Add "a" to get ["a", "a"].
  - Recurse with start = 2.
- Check substring "ab" (indices 1 to 2): Not a palindrome.
Third Level (for partition ["a", "a"], start = 2):
- Check substring "b" (indices 2 to 2): It is a palindrome.
  - Add "b" to get ["a", "a", "b"].
  - Since start becomes 3 (end of string), add this partition to the result.
Backtracking:
- Backtrack and try the partition starting with "aa" from the first level:
  - For partition ["aa"] (start = 2):
    - Check substring "b" (indices 2 to 2): It is a palindrome.
      - Add "b" → partition becomes ["aa", "b"].
      - Recurse with start = 3 and add this partition to the result.

Final Partitions:
The result will contain:

[["a", "a", "b"], ["aa", "b"]]

A visual tree of recursion would look like:

          ""
         /   \
       "a"   "aa"
       /       \
    "a"         "b"
    /             \
  "b"             (end)
   (end)

Common Mistakes, Edge Cases & Alternative Variations

Common Mistakes:

Incorrect Palindrome Check:
Failing to correctly check if a substring is a palindrome (e.g., not handling even-length palindromes properly).
Inefficient Backtracking:
Not backtracking correctly by forgetting to remove the last element, which can cause incorrect partitions.
Missing Base Case:
Not handling the case when the entire string has been partitioned, leading to incomplete or incorrect results.

Edge Cases:

Single Character String:
The string "a" should return [["a"]].
All Characters Same:
For a string like "aaa", there are multiple valid partitions (e.g., ["a","a","a"], ["aa","a"], ["a","aa"], ["aaa"]).
Empty String:
Although the constraints specify at least one character, always consider what your function should do if the input is empty.

Alternative Variations:

Minimum Palindrome Partitioning:
Find the minimum number of cuts needed to partition the string such that every substring is a palindrome.
Longest Palindromic Substring:
Given a string, find the longest substring which is a palindrome.
Palindrome Partitioning II:
Variations where you may need to count the number of palindromic partitions or return the partition with specific constraints.