<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 isPallin(string st) {
        for (int i = 0; i <= st.size() / 2; ++i) {
            if (st[i] != st[st.size() - i - 1])
                return false;
        }
        return true;
    }
    // int func(int ind, string s, int n, vector<int>& dp) {
    //     if (ind == n) {
    //         return 0;
    //     }
    //     if (dp[ind] != -1) {
    //         return dp[ind];
    //     }
    //     string st = "";
    //     int cost = INT_MAX;
    //     int minCost = cost;
    //     for (int partition = ind; partition <= (n - 1); ++partition) {
    //         st += s[partition];
    //         if (isPallin(st)) {
    //             cost = 1 + func(partition + 1, s, n, dp);
    //             minCost = min(minCost, cost);
    //         }
    //     }
    //     return dp[ind] = minCost;
    // }
    int minCut(string s) {
        int n = s.size();
        vector<int> dp(s.size() + 1, 0);
        vector<vector<bool>> isPalindrome(n, vector<bool>(n, false));
        // return func(0, s, s.size(), dp) -1;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (s[i] == s[j] &&
                    (i - j <= 2 || isPalindrome[j + 1][i - 1])) {
                    isPalindrome[j][i] = true;
                }
            }
        }

        for (int ind = n - 1; ind >= 0; --ind) {
            string st = "";
            int cost = INT_MAX;
            int minCost = cost;
            for (int partition = ind; partition <= (n - 1); ++partition) {
                st += s[partition];
                if (isPalindrome[ind][partition]) {
                    cost = 1 + dp[partition + 1];
                    minCost = min(minCost, cost);
                }
            }
            dp[ind] = minCost;
        }
        return dp[0] - 1;
    }
};
33338