<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 t[2001][2001];
    bool palindromeTable[2001][2001];
    
    // Preprocess to check if substring s[i..j] is a palindrome
    void preprocessPalindrome(string &s) {
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            palindromeTable[i][i] = true; // Single characters are palindromes
        }
        
        for (int i = 0; i < n - 1; ++i) {
            palindromeTable[i][i + 1] = (s[i] == s[i + 1]); // Check for two character palindromes
        }
        
        for (int length = 3; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                palindromeTable[i][j] = (s[i] == s[j] && palindromeTable[i + 1][j - 1]);
            }
        }
    }
    
    int solve(string &s, int i, int j) {
        if (i >= j) return 0; // No cuts needed if it's a single character or empty
        if (palindromeTable[i][j]) return 0; // No cut needed if it's already a palindrome
        if (t[i][j] != -1) return t[i][j]; // Memoized value
        
        int ans = INT_MAX;
        for (int k = i; k < j; ++k) {
            if (palindromeTable[i][k]) { // Only proceed if s[i..k] is a palindrome
                int right = solve(s, k + 1, j);
                ans = min(ans, 1 + right); // 1 cut for s[i..k] + solution for s[k+1..j]
            }
        }
        
        return t[i][j] = ans;
    }
    
    int minCut(string s) {
        memset(t, -1, sizeof(t));
        memset(palindromeTable, false, sizeof(palindromeTable));
        preprocessPalindrome(s);
        
        return solve(s, 0, s.size() - 1);
    }
};

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