<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:
    void generatePartitions(string &s, vector<string> &currentPartition, int start, vector<vector<string>> &allPartitions) {
    if (start == s.length()) {
        allPartitions.push_back(currentPartition);
        return;
    }

    for (int i = start; i < s.length(); ++i) {
        currentPartition.push_back(s.substr(start, i - start + 1));
        generatePartitions(s, currentPartition, i + 1, allPartitions);
        currentPartition.pop_back();  // Backtrack
    }
    }
    bool check(string chk)
    {
        int start = 0;
        int end = chk.length()-1;
        while(start<end)
        {
            if(chk[start] != chk[end])
            {
                return false;
            }
            start++;
            end--;
        }
        return true;
    }
    bool isPalindrome(vector<string> vs)
    {
        for(int i = 0 ; i < vs.size() ; i++)
        {
            if(!check(vs[i]))
            {
                return false;
            }
        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        vector<string> currentPartition;
        vector<vector<string>> allPartitions;
        vector<vector<string>> result;
        generatePartitions(s,currentPartition,0,allPartitions);
        for(int i = 0 ; i < allPartitions.size() ; i++)
        {
            if(isPalindrome(allPartitions[i]))
            {
                result.push_back(allPartitions[i]);
            }
        }
        return result;
    }
};
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