<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:
    int dp[2048][2048];
    bool checkPartitioning(string s) {
        int n=s.size();
        for(int i=n-1;i>=0;i--){
            for(int j=i;j<n;j++){
                if(i==j){
                    dp[i][j]=1;
                }else if(j==i+1){
                    dp[i][j]=(s[i]==s[j]?1:0);
                }else{
                    if(s[i]==s[j] && dp[i+1][j-1]){
                        dp[i][j]=1;
                    }else{
                        dp[i][j]=0;
                    }
                }
            }
        }


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