<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 solve(string &s, vector<int> &dp, int i, vector<vector<bool>> &palin){
        int n = s.size();
        int j = n-1;
        if(i>=j) return 0;

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

        if(palin[i][n-1]) return dp[i]=0;

        int mini = 1e9;
        for(int k=i;k<=j;k++){
            if(palin[i][k]){
                int r = solve(s, dp, k+1, palin);
                if(r!=1e9) mini = min(mini, 1+r);
            }
        }

        if(mini==1e9) return dp[i]=1e9;

        return dp[i] = mini;

    }

    int minCut(string s) {
        int n = s.size();
        vector<int> dp(n, -1);
        vector<vector<bool>> palin(n, vector<bool>(n));

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (s[i] == s[j] && (i - j <= 2 || palin[j + 1][i - 1])) {
                    palin[j][i] = true;
                }
            }
        }

        dp[0] = solve(s, dp, 0, palin);
        return dp[0];
    }
};
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