<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 str) {
        int n = str.size();
        vector<vector<bool>> dp(n,vector<bool>(n,false));

        for(int i=n-1;i>=0;i--){
            for(int j=i;j<n;j++){
                int len  = j-i+1;
                if(len==1)dp[i][j]=true;
                else if(len==2)dp[i][j]=(str[i]==str[j]);
                else dp[i][j] = (str[i]==str[j])and(dp[i+1][j-1]);
            }
        }
  
    //now we take (p,s), two non-empty places to cut.
    for(int p=0;p<n;p++){
        for(int s=p+2;s<n;s++){//non-empty
            auto left = dp[0][p];    //if 0 to p is palindrome
            auto mid = dp[p+1][s-1]; //if p+1 to s-1 is palindrome
            auto right = dp[s][n-1]; //if this is palindrome
            if(left && right && mid)return true;
        }
    }
    return false;
    }
};
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