<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 isPalindrome(string& s, int start, int end) {
        int i = start, j = end;
        while (start <= end) {
            if (s[start] != s[end]) {
                return false;
            }
            start++, end--;
        }
        return true;
    }
    int solve(int start, int end, string& s, vector<vector<int>>& dp) {
        if (isPalindrome(s, start, end)) {
            return 0;
        }

        if (dp[start][end] != -1) {
            return dp[start][end];
        }

        int ans = 1e9;
        for (int idx = start; idx < end; idx++) {
            int temp = 1e9;
            
            if(isPalindrome(s, start, idx)) {
                temp = 1 + solve(idx + 1, end, s, dp);
            }

            ans = min(ans, temp);
        }

        return dp[start][end] = ans;
    }

public:
    int minCut(string s) {
        int n = s.length();
        vector<vector<int>> dp(n, vector<int>(n, -1));
        return solve(0, n - 1, s, dp);
    }
};
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