<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:
    vector<vector<int>> isPalindrome;

    void preProcess(string s) {
        int n = s.size();
        isPalindrome.resize(n, vector<int>(n, 0));
        for (int i = 0; i < n; i++) {
            isPalindrome[i][i] = 1;
            if (i + 1 < n && s[i] == s[i + 1]) isPalindrome[i][i + 1] = 1;
        }
        for (int len = 3; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                if (isPalindrome[i + 1][j - 1] == 1 && s[i] == s[j]) {
                    isPalindrome[i][j] = 1;
                }
            }
        }
    }

    bool checkPartitioning(string s) {
        preProcess(s);
        int n = s.size();
        for (int i = 1; i < n - 1; i++) {
            for (int j = i; j < n - 1; j++) {
                if (isPalindrome[0][i - 1] && isPalindrome[i][j] && isPalindrome[j + 1][n - 1]) {
                    // Print the three palindromic substrings
                    cout << s.substr(0, i) << " " << s.substr(i, j - i + 1) << " " << s.substr(j + 1 ) << endl;
                    return true;
                }
            }
        }
        return false;
    }
};

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