<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 {
// #include <iostream>
// #include <vector>
// #include <deque>
// #include <string>
// #include <algorithm>
// #include <limits>

// using namespace std;

public:
    int func(const string& word) {
        int n = word.length();
        vector<int> dp(n + 1, -1);
        int N = n - 1;

        // BASE CASE
        dp[n] = 0;

        for (int i = n - 1; i >= 0; --i) {
            deque<char> q1;
            deque<char> q2;
            dp[i] = numeric_limits<int>::max();
            for (int k = i; k <= N; ++k) {
                q1.push_back(word[k]);
                q2.push_front(word[k]);
                if (q1 == q2) {
                    dp[i] = min((k != N ? 1 : 0) + dp[k + 1], dp[i]);
                }
            }
        }

        return dp[0];
    }

    int minCut(string s) {
        return func(s);
    }
};
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