<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 n = s.size();
        for (int i = 0; i <= n / 2; i++) {
            if (s[i] != s[n - 1 - i]) {
                return 0;
            }
        }
        return 1;
    }

    vector<vector<string>> partition(string s) {
        int n = s.size();
        vector<vector<string>> result[n + 1];
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (isPalindrome(s.substr(j, i - j))) {
                    if(j == 0)
                    {
                        result[i].push_back({s.substr(0, i)});
                    }
                    for (auto it : result[j]) {
                        auto cur = it;
                        cur.push_back(s.substr(j, i-j));
                        result[i].push_back(cur);
                    }
                }
            }
        }
        return result[n];
    }
};
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