<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 is_pal(string s){
        if(s.size() < 2) return true;

        for(int i = 0, j = s.size() - 1; i < j; i++, j--){
            if(s[i] != s[j]) return false;
        } 
        return true;
    }

    void solve(string str, vector<vector<string>>& res, vector<string> ans, int idx){
        if(idx == str.size()){
            res.push_back(ans);
        }

        for(int i = idx; i <= str.size(); i++){
            string sstr = str.substr(idx, i - idx + 1);
            if(is_pal(sstr)){
                ans.push_back(sstr);
                solve(str, res, ans, i + 1);
                ans.pop_back();
            }
        }
    }



    vector<vector<string>> partition(string s) {
        vector<vector<string>> res;
        solve(s, res, {}, 0);
        return res;
    }
};
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