<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 sub){
        if(sub.size()==1){
            return true;
        }
        stack<char> st;
        for(int i=0;i<sub.size();i++){
            st.push(sub[i]);
        }
        string temp="";
        while(!st.empty()){
            temp+=st.top();
            st.pop();
        }

        if(temp==sub){
            return true;
        }
        return false;
    }
    void solve(string s, vector<vector<string>> &ans, vector<string> &temp,
    int ind){
        if(ind==s.size()){
            ans.push_back(temp);
            return;
        }

        for(int i=ind;i<s.size();i++){
            string sub=s.substr(ind, i-ind+1);
            if(isPalindrome(sub)){
                temp.push_back(sub);
                solve(s, ans, temp, i+1);
                temp.pop_back();
            }
        }
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ans;
        vector<string> temp;
        solve(s, ans, temp, 0);

        return ans;
    }
};
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