<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 = (int)s.size();

        int dp[n][n];
        int isPal[n][n];
        for (int i = 0; i < n; ++i) {
            isPal[i][i] = 1;
            dp[i][i] = 0;
        }

        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    if (i + 1 > j - 1)
                        isPal[i][j] = 1;
                    else
                        isPal[i][j] = isPal[i + 1][j - 1];
                    if (isPal[i][j]) {
                        dp[i][j] = 0;
                    }
                } else
                    isPal[i][j] = 0;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            if(isPal[i][n-1]) continue;
            else{
                int ans=INT_MAX;
                for(int j=i;j<n;++j){
                    if(isPal[i][j]){
                        ans= min(ans, 1+ (j+1 < n ? dp[j+1][n-1] : 0));
                    }
                }
                if(ans==INT_MAX) ans= n-i-1;
                dp[i][n-1]=ans;
            }
        }
        return dp[0][n - 1];
    }
};
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