<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string">palindrome</span>.</p>

<p>Return <em>the <strong>minimum</strong> cuts needed for a palindrome partitioning of</em> <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aab&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The palindrome partitioning [&quot;aa&quot;,&quot;b&quot;] could be produced using 1 cut.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ab&quot;
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> isPalindrome(n, vector<bool>(n, false));
        vector<vector<int>> minCuts(n, vector<int>(n, 0));

        // Initialize the isPalindrome table
        for (int length = 1; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                if (s[i] == s[j] && (length <= 2 || isPalindrome[i + 1][j - 1])) {
                    isPalindrome[i][j] = true;
                }
            }
        }

        // Initialize the minCuts table
        for (int i = 0; i < n; ++i) {
            if (isPalindrome[0][i]) {
                minCuts[0][i] = 0;
            } else {
                minCuts[0][i] = INT_MAX;
                for (int j = 0; j < i; ++j) {
                    if (isPalindrome[j + 1][i]) {
                        minCuts[0][i] = min(minCuts[0][i], minCuts[0][j] + 1);
                    }
                }
            }
        }

        return minCuts[0][n - 1];
    }
};

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