<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,vector<vector<string>>> cache;

bool isPalindrome(string s)
{
    int i = 0, j = s.length()-1;
    while(i<j)
    {
        if(s[i]!=s[j]) return false;
        i++;
        j--;
    }
    return true;
}

vector<vector<string>> proc(string s)
{
    if(cache[s].size()!=0) return cache[s];
    if(s.length() == 0) return {{}};
    if(s.length() == 1) cache[s] = {{s}};
    else for(int i = 1;i<=s.length();i++)
    {
        string cur = s.substr(0,i);
        if(isPalindrome(cur))
        {
            // alpha.push_back(cur);
            //cout<<cur<<endl;
            vector<vector<string>> temp;
            temp = proc(s.substr(i));
            for(auto it : temp)
            {
                reverse(it.begin(),it.end());
                it.push_back(cur);
                reverse(it.begin(),it.end());
                cache[s].push_back(it);
            }
            // alpha.pop_back();
        }
    }
    return cache[s];
}

vector<vector<string>> partition(string s) 
{
    vector<vector<string>> retval; vector<string> alpha;
    retval = proc(s);
    return retval;
}
};
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