<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 ispal(string& s,int i,int j){
        while(i<j){
            if(s[i++]!=s[j--]) return 0;
        }
        return 1;
    }
    bool checkPartitioning(string s) {
       int n = s.length(),i,j;
       vector<vector<int>>dp(n,vector<int>(n,-1));
       for(int i = 0;i<n;i++) dp[i][i] = true;
       for(int i = 0;i<=n - 3;i++){
            for(int j = i + 1;j<=n - 2;j++){
                // cout << "i " << i << " j " << j << endl;
                if(dp[0][i]==-1) dp[0][i] = ispal(s,0,i);
                if(dp[i + 1][j]==-1) dp[i + 1][j] = ispal(s,i + 1,j);
                if(dp[j + 1][n - 1]== -1) dp[j + 1][n - 1] = ispal(s,j + 1,n - 1);
                if(dp[0][i] && dp[i + 1][j] && dp[j + 1][n - 1]) {
                    // cout << "i " << i << " j " << j << endl;
                    return true;
                }
            }
       }
       return false;
    }
};
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