<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 isPalin(int i , int j , string& str){
        while(i<j){
            if(str[i]!=str[j]) return false;
            i++;
            j--;
        }
        return true;
    }
    bool solve(int i , int j , int cuts , string& s){
        if(i==j){
            if(cuts!=0) return false;
            return true;
        }
        if(cuts==1)
        {
            if(isPalin(i,j-1,s)) return true;
            else return false;
        }
        bool ans=false;
        for(int index=i;index<j;index++){
           ans|=(isPalin(i,index,s) && solve(index+1,j,cuts-1,s));
        }
        return ans;
    }
    bool checkPartitioning(string s) {
        int j=s.size();
        return solve(0,j,3,s);
    }
};
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