<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 a) {
        int i = 0, j = a.length() - 1;
        while(i < j) {
            if (a[i] != a[j]) {
                return false;
            }
            i++; j --;
        }
        return true;
    }

    void genPalSub(vector<vector<string>> &res, vector<string> curr, string val, int ptr) {
        if (ptr == val.length()) {
            res.push_back(curr);
            return;
        }
        vector<string> currVal = curr; 
        for (int i = 1; i <= val.length() - ptr; i++) {
            string currSub = val.substr(ptr, i);
            if (isPalindrome(currSub)) {
                currVal.push_back(currSub);
                genPalSub(res, currVal, val, ptr + i);
                currVal = curr;
            }
        }
    }
    vector<vector<string>> partition(string s) {
        int n = s.length();
        vector<vector<string>> result;
        vector<string> curr(0);
        
        genPalSub(result, curr, s, 0);
        return result;
    }
};
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