<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 dp[2001][2001];
    int solve(int left,int right,vector<vector<int>>&pal){
        if(left>=right) return 0;
        if(dp[left][right]!=-1) return dp[left][right];
        if(pal[left][right]) return dp[left][right]=0;
        int ans=INT_MAX;
        for(int k=left;k<right;k++){
            if(pal[left][k]) ans=min(ans,1+solve(k+1,right,pal));
        }
        return dp[left][right]=ans;
    }
    int minCut(string s) {
        int n=s.size();
        memset(dp,-1,sizeof(dp));
        vector<vector<int>>pal(n,vector<int>(n,0));
        for(int i=0;i<n;i++) pal[i][i]=1;
        for(int i=0;i<n-1;i++) if(s[i]==s[i+1]) pal[i][i+1]=1;
        for(int i=2;i<n;i++){
            for(int j=0;j+i<n;j++){
                if(s[j]==s[j+i]) pal[j][j+i]=pal[j+1][j+i-1];
            }
        }
        return solve(0,n-1,pal);
    }
};
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