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.

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.

Code

Time Complexity: $O(2^n)$, Space Complexity: $O(n)$

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        if(s.empty()) return res;
        vector<string> cur;
        dfs(0, s, cur, res);
        return res;
    }
 
    void dfs(int index, string& s, vector<string>& cur, vector<vector<string>>& res) {
        if(index == s.length()) {
            res.push_back(cur);
            return;
        }
        for(int i = index; i < s.length(); i++) {
            if(isPalindrome(s, index, i)) {
                cur.push_back(s.substr(index, i - index + 1));
                dfs(i + 1, s, cur, res);
                cur.pop_back();
            }
        }
    }
 
    bool isPalindrome(const string& s, int l, int r) {
        while(l <= r) {
            if(s[l++] != s[r--]) 
                return false;
        }
        return true;
    }
};

Source

• Palindrome Partitioning - LeetCode