<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<vector<double>> map;
        vector<int> minCuts;
        map.push_back({0});
        minCuts.push_back(0);
        for(int i = 1; i < s.length(); i++) {
            int maxCut = minCuts[minCuts.size() - 1] + 1;
            if(s[i] == s[i - 1]) {
                map.push_back({(double)(i + i - 1) / 2, (double)i});
                if(i - 2 >= 0) maxCut = minCuts[i - 2] + 1;
                else maxCut = 0;
            }
            else map.push_back({(double)i});
            for(double center: map[i - 1]) {
                int start = center - (i - center); // check if the same center still works for i
                if(start >= 0 && s[start] == s[i]){
                    map[i].push_back(center);
                    if(start - 1 >= 0) maxCut = min(maxCut, minCuts[start - 1] + 1);
                    else maxCut = 0;
                }
            }
            minCuts.push_back(min(maxCut, minCuts[minCuts.size() - 1] + 1));
        }
        
        return minCuts[s.length() - 1];
    }
};
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