<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:
    void solve(int n, string s, int id, vector<string> &res, vector<vector<string>> &ans, vector<vector<bool>> dp){
        if(id==n){
            ans.push_back(res);
        }
        for(int i=id; i<n; i++){
            string just = s.substr(id, i-id+1);
            if(dp[id][i]){
                res.push_back(just);
                solve(n, s, i+1, res, ans, dp);
                res.pop_back();
            }
        }
    } 
    vector<vector<string>> partition(string s) {
        int n = s.size();
        vector<vector<bool>> dp(n, vector<bool>(n, false));
        for(int i=0; i<n; i++){
            dp[i][i] = true;
        }
        for(int l=2; l<=n; l++){
            for(int i=0; i<=n-l; i++){
                int j = i+l-1;
                if((l==2 || dp[i+1][j-1]==true) && s[i]==s[j]){
                    dp[i][j] = true;
                }
            }
        }
        vector<vector<string>> ans;
        vector<string> res;
        solve(s.size(), s, 0, res, ans, dp);
        return ans;
    }
};
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