<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 {

	bool dp[2005][2005];
	bool vis[2005][2005];
	int n;

	bool isPal(int i, int j, string &s) {
		if (i >= j) {
			return true;
		}
		if (vis[i][j]) {
			return dp[i][j];
		}
		vis[i][j] = 1;
		if (s[i] != s[j]) {
			return dp[i][j] = 0;
		}
		return dp[i][j] = isPal(i + 1, j - 1, s);
	}


public:
    int minCut(string s) {
    	n = s.length();
    	for (int i = 0; i < n; i++) {
    		for (int j = 0; j < n; j++) {
    			vis[i][j] = 0;
    		}
    	}	   
    	if (isPal(0, n - 1, s)) {
    		return 0;
    	}
    	int dp[n];
    	dp[0] = 0;
    	for (int i = 1; i < n; i++) {
    		dp[i] = n + 5;
    		if (isPal(0, i, s)) {
    			dp[i] = 0;
    			continue;
    		}	
    		for (int j = 1; j <= i; j++) {
    			if (isPal(j, i, s)) {
    				dp[i] = min(dp[i], 1 + dp[j - 1]); // j must be greater than or eq to 1
    			}
    		}
    	}
    	return dp[n - 1];
    }
};

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