<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> > palinLen;

    void palindromeCount(const string& s, int left, int right) {
        while (left >= 0 && right < s.length() && s[left] == s[right]) {
            palinLen[left].push_back(right);
            --left;
            ++right;
        }
        return;
    }

    vector<int> dp;

    int solve(int index, string &s)
    {
        int l = s.length();
        if(index==l) return -1;
        if(dp[index]!=-1) return dp[index];

        int ans=1e9;

        for(int x:palinLen[index])
        {
            ans=min(ans,1+solve(x+1,s));
        }

        return dp[index]=ans;
    }

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

        palinLen.assign(l, vector<int>());
        dp.assign(l+1,-1);

        for(int i=0; i<l; i++) palinLen[i].reserve(l);

        for(int i=0; i<l; i++)
        {
            palindromeCount(s,i,i+1);
            palindromeCount(s,i,i);
        }

        int ans = solve(0,s);
        return ans;
    }
};
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