<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<bool>> p_hai;
    int n;
    vector<int> dp;
    int solve(int prev,string& s){
        if(prev==n-1){
            return 0;
        }
        if(dp[prev]!=-1){
            return dp[prev];
        }

        int ans=1e9;
        for(int i=prev;i<n;i++){
            if(p_hai[prev][i]){
                if(i==n-1){
                    ans=min(ans,0);
                }else{
                    ans=min(ans,1+solve(i+1,s));
                }
            }
        }

        return dp[prev]= ans;
    }

    int minCut(string s) {
        n=s.length();

        p_hai.resize(n,vector<bool>(n,true));
        dp.resize(n+1,-1);

        for(int j=1;j<n;j++){
            for(int i=0;i<j;i++){
                p_hai[i][j]= (s[i]==s[j] and (p_hai[i+1][j-1]));
            }
        }

        return solve(0,s);
    }
};
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