<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:
bool pal(string s){
    int i = 0;
    int j = s.length()-1;
    while (i<j){
        if (s[i]!=s[j]){
            return false;
        }
        i++;
        j--;
    }
    return true;
}
    vector<vector<string>> partition(string s) {
        int n = s.length();
       vector<vector<vector<string>>>pi(n);
       vector<string>aux1;
       vector<vector<string>>aux2;
       aux1.push_back(s.substr(0,1));
       aux2.push_back(aux1);
       pi[0]=aux2;
       for (int i = 1; i<n; i++){
        aux2.clear();
        for (int j = 0; j<=i; j++){
            if (pal(s.substr(j,i-j+1))){
                if (j!=0){
                    for (auto e: pi[j-1]){
                        aux1 = e;
                        aux1.push_back(s.substr(j, i-j+1));
                        aux2.push_back(aux1);
                    }
                }
                else {
                    aux1.clear();
                    aux1.push_back(s.substr(j, i-j+1));
                    aux2.push_back(aux1);
                }
            }
        }
        pi[i]=aux2;
       }
       return pi[n-1];
    }
};
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