<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) {
        int n = s.size();
        //length , start index
        vector<vector<int>> dp(n+1,vector<int>(n+1,0));
        for(int i = 0; i<=n; i++){
            for(int j = 0; j<=n-i; j++){
                if(i == 0 || i == 1){
                    dp[i][j] = 1;
                    continue;
                }
                if(s[j] == s[j+i-1] && dp[i-2][j+1] == 1){
                    dp[i][j] = 1;
                }else{
                    dp[i][j] = 0;
                }
            }
        }
        vector<int> jps(n,INT_MAX);
        for(int i = n-1; i>=0; i--){
            if(dp[s.size()-i][i] == 1){
                jps[i] = 0;
                continue;
            }
            for(int j = 1; (i+j)<s.size(); j++){
                if(dp[j][i] == 1){
                    jps[i] = min(jps[i],jps[i+j]+1);
                }
            }
        }
        return jps[0];
    }
};
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