<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:
    bool isPalindrome(int i, int j, string &s){
        while(i<j){
            if(s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }
    int f(int i, int n, string &s, int dp[]){
        if(i==n) return 0;
        if(dp[i] != -1) return dp[i];
        int minCost = INT_MAX;
        for(int j=i; j<n; j++){
            if(isPalindrome(i, j, s)){
                int cost = 1 + f(j+1, n, s, dp);
                minCost = min(minCost, cost);
            }
        }
        return dp[i] = minCost;

    }
    int minCut(string s) {
        int n = s.size();
        int dp[n+1];
        memset(dp, 0, sizeof(dp));
        for(int i=n-1; i>=0; i--){
            int minCost = INT_MAX;
            for(int j=i; j<n; j++){
                if(isPalindrome(i, j, s)){
                    int cost = 1 + dp[j+1];
                    minCost = min(minCost, cost);
                }
            }
            dp[i] = minCost;
        }
        return dp[0]-1;//A|B|C| for this last partition done we are doing -1
    }
};
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