<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:
    vector<vector<int>> dp;
    int mem[2005];
    int solve(int i,string &s){
        int n=s.length();
        if(i==n)return 0;
        if(mem[i]!=-1)return mem[i];
        int ans=1e9;
        for(int k=i;k<n;k++){
            if(dp[i][k]){
                if(k!=n-1){
                    ans=min(ans,1+solve(k+1,s));
                }
                else ans=0;
            }
        }
        return mem[i]=ans;
    }

    int minCut(string s) {
        int n=s.length();
        dp.resize(n,vector<int>(n,0));
        memset(mem,-1,sizeof(mem));
        for(int i=0;i<n;i++){
            dp[i][i]=1;
        }
        for(int i=0;i<n-1;i++){
            if(s[i]==s[i+1]){
                dp[i][i+1]=1;
            }
        }
        for(int length=3;length<=n;length++){
            for(int i=0;i<=n-length;i++){
                if(s[i]==s[i+length-1])dp[i][i+length-1]=dp[i+1][i+length-2];
            }
        }
        return solve(0,s);
    }
};
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