<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:
    unordered_map<int,int>dp;
    unordered_map<int,vector<int>>tap;
    int recur(string &s,int i){
        if(s.length()==i+1)return dp[i]=0;
        if(dp.count(i))return dp[i];
        int result=INT_MAX;int n = s.length();
        for(int j =0;j<tap[i].size();j++){
            int a = tap[i][j];
            if(a==n-i)return dp[i]=0;
            if(a>n-i)continue;
            result=min(result,recur(s,i+a));
        }
        return dp[i]=1+result;
    }
    int minCut(string s) {int n =s.length();
        for(int i = 0;i<n;i++){
            tap[i].push_back(1);
            int left,right;left = i-1;right=i+1;
            while(left>=0&&right<n&&s[left]==s[right]){
                tap[left].push_back(right-left+1);
                left--;right++;
            };
            left=i;right=i+1;
            while(left>=0&&right<n&&s[left]==s[right]){
                tap[left].push_back(right-left+1);
                left--;right++;
            };
        };
        return recur(s,0);
    }
};
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