<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string"><strong>palindrome</strong></span>. Return <em>all possible palindrome partitioning of </em><code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> [["a","a","b"],["aa","b"]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a"
<strong>Output:</strong> [["a"]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 16</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    unordered_map <string,int> mpp;
    bool palin(string& s){
        int i = 0;
        int j = s.size() -1;
        while(i < j){
            if(s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
    bool checkpalin(vector<string>& curr){
        for(int i = 0;i < curr.size();i++){
            if(mpp.find(curr[i]) != mpp.end()){
                if(mpp[curr[i]] == 0) return false;
            }else{
                if(palin(curr[i])){
                    mpp[curr[i]] = 1;
                }else{
                    mpp[curr[i]] = 0;
                    return false;
                }
            }
        }
        return true;
    }
    void f(int index,vector<string>& curr,vector<vector<string>>& ans,string& s,int n){
        if(index == n){
            if(checkpalin(curr))
                ans.push_back(curr);
            return;
        }

        curr.push_back(string(1,s[index]));
        f(index+1,curr,ans,s,n);
        curr.pop_back();

        if(!curr.empty()){
            string temp = curr.back();
            curr.back() += string(1,s[index]);
            f(index+1,curr,ans,s,n);
            curr.back() = temp;
        }
    }

    vector<vector<string>> partition(string s) {
        vector<string> curr;
        vector<vector<string>> ans;
        int n = s.size();
        // return {{}};
        f(0,curr,ans,s,n);
        return ans;        
    }
};
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