<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:
    long long M = 1000000007;
    long long hash(long long ex, char x)
    {
        ex *= 31;
        ex%=M;
        ex += x - 'a' + 1;
        return ex%M;
    }

    bool checkPartitioning(string s) {

        
        long long fwd[s.size()][s.size()];
        long long bwd[s.size()][s.size()];
        memset(fwd, 0, sizeof(fwd));
        memset(bwd, 0, sizeof(bwd));
        for(int i=0; i<s.size(); i++)
        {
            long long h = 0;
            for(int j = i; j<s.size(); j++)
            {
                h = hash(h, s[j]);
                fwd[i][j] = h;
                
            }
        }
        for(int i=s.size()-1; i>=0; i--)
        {
            long long h = 0;
            for(int j = i; j>=0; j--)
            {
                h = hash(h, s[j]);
                bwd[j][i] = h;
            }
        }

        for(int i=0; i<s.size()-2; i++)
        {
            for(int j = s.size()-1; j > 1; j--)
            {
                if( i + 1 > j - 1) break;
                if(fwd[0][i] == bwd[0][i] && fwd[j][s.size()-1] == bwd[j][s.size()-1] && fwd[i+1][j-1] == bwd[i+1][j-1])
                {
                    return true;
                }
            }
        }
        return false;

        
    }
};
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