<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>

using namespace std;

class Solution {
public:
    bool checkPartitioning(string s) {
        int n = s.size();

        // Step 1: Create a 2D array to store whether substrings are palindromes
        vector<vector<bool>> is_palindrome(n, vector<bool>(n, false));

        // Step 2: Initialize the DP table
        // Single characters are palindromes
        for (int i = 0; i < n; ++i) {
            is_palindrome[i][i] = true;
        }
        // Two consecutive characters are palindromes if both are the same
        for (int i = 0; i < n - 1; ++i) {
            is_palindrome[i][i + 1] = (s[i] == s[i + 1]);
        }
        // Fill the DP table for substrings longer than 2 characters
        for (int length = 3; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                is_palindrome[i][j] = (s[i] == s[j]) && is_palindrome[i + 1][j - 1];
            }
        }

        // Step 3: Try to partition the string into 3 palindromic substrings
        for (int i = 1; i < n - 1; ++i) { // First partition point
            for (int j = i + 1; j < n; ++j) { // Second partition point
                if (is_palindrome[0][i - 1] && is_palindrome[i][j - 1] && is_palindrome[j][n - 1]) {
                    return true;  // Found a valid partition
                }
            }
        }

        return false;  // No valid partition found
    }
};
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