<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) {
        vector<int> cutsDp;
        cutsDp.resize(s.length());
        for (int i = 1; i < s.length(); i++) {
            cutsDp[i] = i;
        }
        for (int mid = 0; mid < s.length(); mid++) {
            // check for odd length palindrome around mid index
            findMinimumCuts(mid, mid, cutsDp, s);
            // check for even length palindrome around mid index
            findMinimumCuts(mid - 1, mid, cutsDp, s);
        }
        return cutsDp[s.length() - 1];
    }

    void findMinimumCuts(int startIndex, int endIndex, vector<int> &cutsDp,
                         string s) {
        for (int start = startIndex, end = endIndex;
             start >= 0 && end < s.length() && s[start] == s[end];
             start--, end++) {
            int newCut = start == 0 ? 0 : cutsDp[start - 1] + 1;
            cutsDp[end] = min(cutsDp[end], newCut);
        }
    }
};
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