<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[2001][2001][3];
bool check(string&s, int i, int j){
   while(i<j) if(s[i++]!=s[j--]) return false;
   
   return true;
}

int solve(string&s, int i, int j,int ct){

     if(i>j) return false;
    if(ct==2) {
         if(check(s,i,j)) return true;
         return false;
    }
    if(dp[i][j][ct]!=-1) return dp[i][j][ct];

    for(int k=i; k<j; k++){
         if(check(s, i,k))  if(solve(s, k+1,j,ct+1)) return dp[i][j][ct]=true;
    }
    return dp[i][j][ct]= false;
}

    bool checkPartitioning(string s) {
        memset(dp,-1, sizeof dp);
        return solve(s, 0,s.size()-1,0);
    }
};
49998