<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:
    // Function to check if a substring s[i...j] is a palindrome
    bool isPal(const string &s, int i, int j) {
        while (i < j) {
            if (s[i++] != s[j--]) {
                return false;
            }
        }
        return true;
    }

    bool checkPartitioning(const string &s) {
        int n = s.size();
        // Iterate over possible end points for the third palindrome
        for (int i = 2; i < n; ++i) {
            // Check if s[i...n-1] is a palindrome
            if (isPal(s, i, n - 1)) {
                // Iterate over possible end points for the second palindrome
                for (int j = 1; j < i; ++j) {
                    // Check if s[j...i-1] is a palindrome and s[0...j-1] is a palindrome
                    if (isPal(s, j, i - 1) && isPal(s, 0, j - 1)) {
                        return true; // Found a valid partition into 3 palindromic substrings
                    }
                }
            }
        }
        return false; // No valid partition found
    }
};

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