<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 palindrome[2001][2001];
    int isPalindrome(string& s,int l,int r)
    {
        if(l >= r)
        {return palindrome[l][r] = 1;}
        if(palindrome[l][r] != -1)
        {
            return palindrome[l][r];
        }
        while(l <=r)
        {
            if(s[l] != s[r])
            {
                return palindrome[l][r] = 0;
            }
            l++;
            r--;
        }
        return palindrome[l][r] = 1;
    }
    int minCut(string s) {

        memset(dp,-1,sizeof(dp));
        memset(palindrome,-1,sizeof(palindrome));
        int result = partition(s,0,s.size()-1);
        return result;
    }
    int partition(string& s,int i,int j)
    {
        if(i >= j)
        {
            return 0;
        }
        if(isPalindrome(s,i,j)==1)
        {
            return 0;
        }
        if(dp[i][j] != -1)
        {
            return dp[i][j];
        }
        int ans = INT_MAX;
        for(int k = i;k <= j - 1; k++)
        {
            if(isPalindrome(s,i,k))
            {
            ans = min(ans, 1 + partition(s,k+1,j));
            }
        }
        return dp[i][j] = ans;
    }
};

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