<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 isPalindrome(string s){
    int l = 0, r = s.size()-1;
    while(l <= r){
        if(s[l] != s[r]) return false;
        l++;
        r--;
    }

    return true;
}
void solve(vector<vector<string>> &ans, vector<string> tmp, string s, int n){
    
    for(int i = 0;i < s.size();i++){
        cout << s.substr(0,i+1) << endl;
        if(isPalindrome(s.substr(0,i+1))){
            tmp.push_back(s.substr(0,i+1));
            if(i + 1 < s.size()){
                solve(ans,tmp,s.substr(i+1),n);
            } else{
                ans.push_back(tmp);
            }
            
            if(!tmp.empty()){
                tmp.pop_back();
            }
        }
        
    }
}
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> tmp;
        solve(ans,tmp,s, s.size());
        return ans;
    }
};
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