<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 pal[2005][2005];

    // Helper function to find all palindromes
    void findPals(string &s) {
        int n = s.size();
        memset(pal, false, sizeof(pal));
        
        // Palindrome of single characters
        for (int i = 0; i < n; i++) {
            pal[i][i] = true;  // Single character is always a palindrome
        }

        // Palindrome of two characters
        for (int i = 0; i < n - 1; i++) {
            if (s[i] == s[i + 1]) {
                pal[i][i + 1] = true;
            }
        }

        // Palindromes of length 3 or more
        for (int len = 3; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                int j = i + len - 1;
                if (pal[i + 1][j - 1] && s[i] == s[j]) {
                    pal[i][j] = true;
                }
            }
        }
        return;
    }

    int dp[2005][2005][4];

    bool solve(int ind, int prev, int k, string &s){
        int n = s.size();
        if(k == 1) return pal[prev][n-1];
        if(ind >= n) return false;
        if(dp[ind][prev][k] != -1) return (bool)dp[ind][prev][k];
        
        bool p = false, np = false;
        if(pal[prev][ind]) p = solve(ind+1, ind+1, k-1, s);
        np = solve(ind+1, prev, k, s);

        return dp[ind][prev][k] = p|np;
    }


    bool checkPartitioning(string s) {
        int n = s.size();
        memset(dp, -1, sizeof(dp));
        findPals(s);
        return solve(0, 0, 3, s);

        // for(int i=n-1; i>=0; i--){
        //     for(int j=i; j<n; j++){
        //         for(int k=0; k<3; k++){
                    
        //             dp[i][j][k] = dp[]

        //         }
        //     }
        // }

    }
};
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