<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:
    bool check(int i,int j,string &s,vector<vector<int>> &dp){
        if(i > j) return 1;

        if(dp[i][j] != -1) return dp[i][j];
        if(s[i] == s[j]) return dp[i][j] = check(i+1,j-1,s,dp);
        return dp[i][j] = 0;
    }
    int solve(int i,int n,string &s,vector<vector<int>> &dp1,vector<int> &dp2){
        if(i == n) return -1;
        if(dp2[i] != -1) return dp2[i];
        int count = INT_MAX;
        for(int j = i;j<n;j++){
            if(check(i,j,s,dp1)){
                count = min(count,solve(j+1,n,s,dp1,dp2));
            }
        }
        if(count != INT_MAX) return dp2[i] = 1 + count;
        return dp2[i] = count;
    }
    int minCut(string s) {
        int n = s.length();
        vector<vector<int>> dp1(n,vector<int>(n,-1));
        vector<int> dp2(n+1,-1);
        return solve(0,n,s,dp1,dp2);
    }
};
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