Palindrome Partitioning LeetCode Solution – Queslers

Problem – Palindrome Partitioning

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

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"]]


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

Palindrome Partitioning LeetCode Solution in Java

public List<List<String>> partition(String s) {
        // Backtracking
        // Edge case
        if(s == null || s.length() == 0) return new ArrayList<>();
        List<List<String>> result = new ArrayList<>();
        helper(s, new ArrayList<>(), result);
        return result;
    public void helper(String s, List<String> step, List<List<String>> result) {
        // Base case
        if(s == null || s.length() == 0) {
            result.add(new ArrayList<>(step));
        for(int i = 1; i <= s.length(); i++) {
            String temp = s.substring(0, i);
            if(!isPalindrome(temp)) continue; // only do backtracking when current string is palindrome
            step.add(temp);  // choose
            helper(s.substring(i, s.length()), step, result); // explore
            step.remove(step.size() - 1); // unchoose
    public boolean isPalindrome(String s) {
        int left = 0, right = s.length() - 1;
        while(left <= right) {
            if(s.charAt(left) != s.charAt(right))
                return false;
            left ++;
            right --;
        return true;

Palindrome Partitioning LeetCode Solution in C++

class Solution {
    vector<vector<string>> partition(string s) {
        vector<vector<string> > ret;
        if(s.empty()) return ret;
        vector<string> path;
        dfs(0, s, path, ret);
        return ret;
    void dfs(int index, string& s, vector<string>& path, vector<vector<string> >& ret) {
        if(index == s.size()) {
        for(int i = index; i < s.size(); ++i) {
            if(isPalindrome(s, index, i)) {
                path.push_back(s.substr(index, i - index + 1));
                dfs(i+1, s, path, ret);
    bool isPalindrome(const string& s, int start, int end) {
        while(start <= end) {
            if(s[start++] != s[end--])
                return false;
        return true;

Palindrome Partitioning LeetCode Solution in Python

class Solution(object):
    def __init__(self):
        self.memory = collections.defaultdict(list)
    def partition(self, s):
        if not s: return [[]]
        if s in self.memory: return self.memory[s]  # the memory trick can save some time
        ans = []
        for i in range(1, len(s) + 1):
            if s[:i] == s[:i][::-1]:  # prefix is a palindrome
                for suf in self.partition(s[i:]):  # process suffix recursively
                    ans.append([s[:i]] + suf)
        self.memory[s] = ans
        return ans
Palindrome Partitioning LeetCode Solution Review:

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