<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 minCut(string s) {
        int n=s.size();
        int memo[2000][2000]={0};
        if(n<2){
            return 0;
        }

        vector<int> dp(n+1,n);
        //build the valid Palindrome map
        for(int i=0; i<n; i++){
            int j;
            int k;
            for (j=i,k=i; j>=0&&k<n; j--,k++){
                if(s[j]==s[k]){
                    memo[j][k]=1;
                }else{
                    break;
                }
            }
            for(j=i,k=i+1; j>=0&&k<n; j--,k++){
                if(s[j]==s[k]){
                    memo[j][k]=1;
                }else{
                    break;
                }              
            }
        }


        dp[0]=-1;
        dp[1]=0;

        for(int i=1; i<n; i++){

            for(int j=i; j>=0; j--){

                // string ss=s.substr(j,i-j+1);
                // string rs=ss;
                // reverse(rs.begin(),rs.end());
                string range = to_string(j)+','+to_string(i);
                if( memo[j][i]){
                    dp[i+1] = min(dp[i+1],dp[j]+1);
                }
            }

        }
        return dp[n];
        
    }
};
26539