<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_palindrome(string &s, int st, int en) {
        while(st <= en) {
            if(s[st] != s[en]) {
                return 0;
            }
            st++;
            en--;
        }

        return 1;
    }
    void solve(string &s, int prev, int i, vector<vector<string>> &ans, vector<string> str) {
        if(i == s.length()-1) {
            if(is_palindrome(s, prev, i)) {
                str.push_back(s.substr(prev, i - prev + 1));
                ans.push_back(str);
            }
            return ;
        }

        if(is_palindrome(s, prev, i)) {
            str.push_back(s.substr(prev, i - prev + 1));
            solve(s, i+1, i+1, ans, str);
            str.pop_back();
        }
        solve(s, prev, i+1, ans, str);
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        solve(s, 0, 0, ans, {});

        return ans;
    }
};
247654