<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) {
    const int n = s.length();
    vector<vector<bool>> isPalindrome(n, vector<bool>(n, true));
    vector<int> dp(n, n);

    for (int l = 2; l <= n; ++l)
      for (int i = 0, j = l - 1; j < n; ++i, ++j)
        isPalindrome[i][j] = s[i] == s[j] && isPalindrome[i + 1][j - 1];

    for (int i = 0; i < n; ++i) {
      if (isPalindrome[0][i]) {
        dp[i] = 0;
        continue;
      }

      for (int j = 0; j < i; ++j)
        if (isPalindrome[j + 1][i])
          dp[i] = min(dp[i], dp[j] + 1);
    }

    return dp.back();
  }
};
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