LeetCode-in-All

131. Palindrome Partitioning

Medium

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 string is a string that reads the same backward as forward.

Example 1:

Input: s = “aab”

Output: [[“a”,”a”,”b”],[“aa”,”b”]]

Example 2:

Input: s = “a”

Output: [[“a”]]

Constraints:

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

To solve the “Palindrome Partitioning” problem in Java with a Solution class, we’ll use backtracking. Below are the steps:

1. Create a Solution class: Define a class named Solution to encapsulate our solution methods.

2. Create a partition method: This method takes a string s as input and returns all possible palindrome partitioning of s.

3. Define a recursive helper method: Define a recursive helper method backtrack to find all possible palindrome partitions.
   • The method should take the current index start, the current partition partition, and the list to store all partitions result.
   • Base case: If start reaches the end of the string s, add the current partition to the result list and return.
   • Iterate from start to the end of the string:
     • Check if the substring from start to i is a palindrome.
     • If it is a palindrome, add the substring to the current partition and recursively call the backtrack method with the updated index and partition.
     • After the recursive call, remove the last substring added to the partition to backtrack and explore other partitions.

4. Initialize a list to store all partitions: Create an ArrayList named result to store all possible palindrome partitions.

5. Call the helper method: Call the backtrack method with the initial index, an empty partition list, and the result list.

6. Return the result list: After exploring all possible partitions, return the list containing all palindrome partitions.

Here’s the Java implementation:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> result = new ArrayList<>();
        backtrack(s, 0, new ArrayList<>(), result);
        return result;
    }
    
    // Recursive helper method to find all possible palindrome partitions
    private void backtrack(String s, int start, List<String> partition, List<List<String>> result) {
        if (start == s.length()) {
            result.add(new ArrayList<>(partition));
            return;
        }
        
        for (int i = start; i < s.length(); i++) {
            String substring = s.substring(start, i + 1);
            if (isPalindrome(substring)) {
                partition.add(substring);
                backtrack(s, i + 1, partition, result);
                partition.remove(partition.size() - 1); // Backtrack
            }
        }
    }
    
    // Helper method to check if a string is a palindrome
    private boolean isPalindrome(String s) {
        int left = 0;
        int right = s.length() - 1;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

This implementation follows the steps outlined above and efficiently finds all possible palindrome partitions of the given string in Java.

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