<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 static t[2002][2002];
class Solution {
    
public:
    bool ispalindrome(string &s, int i,int j){
        if(i>=j){
            return true;
        }
        while(i<j){
            if(s[i]!=s[j]){
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
    int solve(string &s,int i,int j){
        if(i>=j){
            return 0;
        }
        if(ispalindrome(s,i,j)){
            return 0;
        }
        if(t[i][j]!=-1){
            return t[i][j];
        }
        int mini=INT_MAX;
        for(int k=i;k<j;k++){
            if(ispalindrome(s, i, k)){                         
                int temp = solve (s, k+1, j) + 1;
                mini = min (mini, temp);
            }
        }
        return t[i][j]=mini;
    }
    int minCut(string s) {
        int n=s.size();
        memset(t,-1,sizeof(t));
        return solve(s,0,n-1);
    }
};
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