<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>

int dp1[2005][2005];
int dp2[2005];

int check(int l, int r, string &s)
{
    if(l>=r)
    return dp1[l][r]=1;

    if(dp1[l][r]!=-1)
    return dp1[l][r];

    if(s[l]==s[r])
    return dp1[l][r]=check(l+1,r-1,s);
    else
    return dp1[l][r]=0;
}

int solve(int i, string &s)
{
    int n=s.size();
    if(i==n)
    return 0;

    if(dp2[i]!=-1)
    return dp2[i];

    int ans=n,j;
    for(j=i;j<n;j++)
    {
        if(check(i,j,s))
        ans=min(ans,1+solve(j+1,s));
    }
    return dp2[i]=ans;
}

class Solution {
public:
    int minCut(string s) {
        memset(dp1,-1,sizeof(dp1));
        memset(dp2,-1,sizeof(dp2));
        return solve(0,s)-1;
    }
};
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