Palindrome Partitioning II
<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>
String; Dynamic Programming
class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        bool g[n][n];
        memset(g, true, sizeof(g));
        for (int i = n - 1; ~i; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s[i] == s[j] && g[i + 1][j - 1];
            }
        }
        int f[n];
        iota(f, f + n, 0);
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (g[j][i]) {
                    f[i] = min(f[i], j ? 1 + f[j - 1] : 0);
                }
            }
        }
        return f[n - 1];
    }
};