<p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.​​​​​</p>

<p>A string is said to be palindrome if it the same string when reversed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcbdd&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&quot;abcbdd&quot; = &quot;a&quot; + &quot;bcb&quot; + &quot;dd&quot;, and all three substrings are palindromes.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bcbddxy&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>s cannot be split into 3 palindromes.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
</ul>

class Solution {

public:
    bool checkPartitioning(string s) {
        int n = s.size();
        if(n == 3) return true;
        vector<vector<int>> dp(n, vector<int>(n, -1));

        // Initialise the results for substrings of length 1 and 2.
        for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
            if(i < n-1) dp[i][i+1] = s[i] == s[i+1];
        }

        for (int len = 3; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                dp[i][i + len - 1] = s[i] == s[i + len - 1] and dp[i+1][i + len - 2];
            }
        }

        for (int i = 1; i < n-1; i++) {
            for (int j = i; j < n-1; j++) {
                if(dp[0][i-1] and dp[i][j] and dp[j+1][n-1]) return true;
            }
        }

        return false;    

    }
};

auto init = []() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();
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