<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 {
    int rec(string S, int st, int end, vector<vector<int>>& dp,
            vector<vector<int>> isPal) {
        if (isPal[st][end])
            return 0;
        else {
            if (dp[st][end] != -1)
                return dp[st][end];
            int ans = INT_MAX;
            for (int i = st; i < end; ++i) {
                if (isPal[st][i]) {
                    int part = 1 + rec(S, i + 1, end, dp, isPal);
                    ans = min(ans, part);
                } else {
                    continue;
                }
            }
            return dp[st][end] = ans;
        }
    }

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 j = 0; j < i; ++j) {
                isPal[i][j] = 1;
                dp[i][j] = 0;
            }
        }

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