<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:
    vector<vector<string>> partition(string s) {
        int n=s.size();
        vector<vector<bool>> t(n,vector<bool>(n,false));
        vector<vector<string>> result;
        vector <string> path;
        
        for(int i=0;i<n;i++)
            t[i][i]=true;
        
        for(int k=2;k<=n;k++){
            for(int i=0;i<n-k+1;i++){
                int j=i+k-1;
                    if(s[i]==s[j]){
                        if(k==2)
                            t[i][j]=true;
                        else
                            t[i][j]=t[i+1][j-1];
                    }
            }
        }

        fun(0,s,result,path,t);
        
        return result;
    }
    
    void fun(int index,string s,vector<vector<string>> &result,vector <string> &path,vector<vector<bool>> t){
        
        if(index==s.size()){
         result.push_back(path);
         return;
        }
        
        for(int i=index;i<s.size();++i){
            if(t[index][i]==true){
                path.push_back(s.substr(index,i-index+1));
                fun(i+1,s,result,path,t);
                path.pop_back();
            }
        } 
    }

};
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