<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 ispallindrome(string s, int i) {

        int n = s.length();
        if (i >= n / 2)
            return true;
        if (s[i] != s[n - 1 - i]) {
            return false;
        }
        return ispallindrome(s, i + 1);
    }

    void sol(vector<vector<string>>& ans, vector<string>& ds, string s,
             string d, int i) {
        if (i == s.length()) {
            for (auto i : ds) {
                if (ispallindrome(i, 0) == false) {
                    return;
                }
            }
            ans.push_back(ds);
            return;
        }
        d += s[i];
        if (i < s.length() - 1) {
            sol(ans, ds, s, d, i + 1);
        }
        ds.push_back(d);
        sol(ans, ds, s, "", i + 1);
        ds.pop_back();
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> ds;
        string d = "";
        int i = 0;
        sol(ans, ds, s, d, i);
        return ans;
    }
};
113831