<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:

    // unordered_map<int, unordered_map<int, int>> map1;

    int check(string &s, int i, int j, vector<vector<int>>&map1){
        while(i < j){
            if(s[i] != s[j]) return map1[i][j] = 2;
            i++;
            j--;
        }
        map1[i][j] = 1;
        return 1;
    }

    bool solve(string &s, int i, int cnt, vector<vector<int>>&dp, vector<vector<int>>&dp1){
        if(i >= s.size() && cnt == 0){
            return true;
        }
        if(i >= s.size() || cnt < 0){
            return false;
        } 
        if(dp[i][cnt] != -1) return dp[i][cnt];
        char c = s[i];
        string s1 = "";
        for(int j=i;j<s.size();j++){
            if(s[j] == c){
                if(dp1[i][j] == 1){
                    if(solve(s, j+1, cnt-1, dp, dp1)) return dp[i][cnt] = true;
                }
            }
        }
        return dp[i][cnt] = false;
    }

    bool checkPartitioning(string s) {
        // map1.clear();
        vector<vector<int>> dp1(s.size(), vector<int>(s.size(), 1));
        for(int i=0;i<s.size();i++){
            for(int j=i+1;j<s.size();j++){
                dp1[i][j] = check(s, i, j, dp1);
            }
        }
        vector<vector<int>> dp(s.size(), vector<int>(4, -1));
        return solve(s, 0, 3, dp, dp1);
    }
};
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