131. Palindrome Partitioning

Description

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: "aab"

Output:

[

["aa", "b"],

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

]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> resultList = new ArrayList();
        if(s == null || s.length() == 0) return resultList;
        helper(resultList, new ArrayList(), s, 0);
        return resultList;
    }
    
    private void helper(List<List<String>> resultList,
                        List<String> currList,
                        String s,
                        int start) {
        if(start == s.length()) {
            resultList.add(new ArrayList(currList));
            return;
        }
        
        for(int i = start; i < s.length(); i++) {
            if(isPalindrome(s, start, i)) {
                currList.add(s.substring(start, i+1));
                helper(resultList, currList, s, i+1);
                currList.remove(currList.size()-1);
            }
        }
    }
    
    private boolean isPalindrome(String s, int left, int right) {
        while(left < right) {
            if(s.charAt(left) != s.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

Follow up