<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:
    static bool checkPalindrome(string currentSubstring,int start,int end){
        if(start>=end) return 1;
        if(currentSubstring[start]!=currentSubstring[end]){
            return 0;
        }
        return checkPalindrome(currentSubstring,start+1,end-1);
    }

    static void recursion(string s, vector<vector<string>>& ans, vector<string>& singleAns, int count) {
        if(count==s.size()){
            ans.push_back(singleAns);
            return;
        }
        
        for (int i = count; i < s.size(); i++) {
            if (checkPalindrome(s, count, i)) {
                singleAns.push_back(s.substr(count, i - count + 1));
                recursion(s, ans, singleAns, i + 1);
                singleAns.pop_back();
            }
        }
    }

    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> singleAns;
        recursion(s,ans,singleAns,0);
        return ans;
    }
};

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