<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 isPal(string &s){
        for(int i=0;i<s.size()/2;i++){
            if(s[i]!=s[s.size()-1-i]) return false;

        }
        return true;
    }
    vector<vector<string>> partition(string s) {
        int n=s.size();
        vector<vector<string>>ans;
        for(int i=0;i<(1LL<<(n-1));i++){
            vector<int>temp;
            for(int j=0;j<n;j++){
                if(i&(1LL<<j)){
                    temp.push_back(j);
                }
            }
            int st=0;
            vector<string>v;
            for(int ii=0;ii<temp.size();ii++){
                v.push_back(s.substr(st,temp[ii]-st+1));
                st=temp[ii]+1;
            }
            if(st<n) v.push_back(s.substr(st,n-st));
            int f=0;
            for(auto it:v){
                if(!isPal(it)){
                    f=1;
                    break;
                }
            }
            if(!f) ans.push_back(v);           

        }
        return ans;
        
    }
};
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