<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:
    Solution() {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
    }
    bool isPalindrome(string& s, int i, int j) {
        while(j < i) {
            if(s[j++] != s[i--])
                return false;
        }
        return true;
    }

    vector<vector<string>> partition(string s) {
        int n = s.size();
        vector<vector<vector<string>>> dp(n + 1, vector<vector<string>>());
        dp[0].emplace_back();
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= i; j++) {
                if(isPalindrome(s, i, j)) {
                    for(int k = 0; k < dp[j].size(); k++) {
                        dp[i + 1].emplace_back(dp[j][k].begin(), dp[j][k].end());
                        dp[i + 1].back().push_back(s.substr(j, i - j + 1));
                    }
                }
            }
        }
        return dp[n];
    }
};
174370