<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.length();
        vector<vector<int>>p(n+1,vector<int>(n+1,0));
        for(int i=0;i<n;i++){
            p[i][i]=1;
            if(i!=n-1 && s[i]==s[i+1]){
                p[i][i+1]=1;
            }
        }
        for(int gap=2;gap<n;gap++){
            for(int i=0;i+gap<n;i++){
                int j=i+gap;
                if(s[i]==s[j] && p[i+1][j-1]){
                    p[i][j]=1;
                }
                else{
                    p[i][j]=0;
                }
            }
        }

        bool ans=false;
        for(int i=0;i<n-2;i++){
            for(int j=i+1;j<n-1;j++){
                if(p[0][i] && p[i+1][j] && p[j+1][n-1]){
                    ans=true;
                    break;
                }
            }
        }
        return ans;
    }
};
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