<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 check(string &s,int st,int end){
        while(st<end){
            if(s[st]!=s[end]) return 0;
            st++,end--;
        }
        return 1;
    }
    int dp[2001][4];
    bool solve(int i,vector<int>v[],int rem,int n){
        if(i==n){
            if(rem==0) return 1;
            return 0;
        }
        if(rem<0) return 0;
        if(dp[i][rem]!=-1) return dp[i][rem];
        bool ans=0;
        for(int j=0;j<v[i].size();j++){
            ans=ans|solve(v[i][j],v,rem-1,n);
        }
        return dp[i][rem]=ans;
    }
    bool checkPartitioning(string s) {
        int n=s.size();
        memset(dp,-1,sizeof(dp));
        vector<int>v[n];
        for(int i=0;i<s.size();i++){
            for(int j=i;j<s.size();j++){
                if(check(s,i,j)){
                    v[i].push_back(j+1);
                }
            }
        }
        return solve(0,v,3,n);
    }
};
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