<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 dp[2005][2005];

    int is_palindrome(string& s) {
        if (s.size() == 1)
            return 1;

        int n = s.size();
        int x = (n + 1) / 2;
        int j = n - 1;
        for (int i = 0; i <= x; i++, j--) {
            if (i > j)
                break;
            if (s[i] != s[j])
                return 0;
        }
        return 1;
    }

    int solve(string& s, int i, int j) {
        if (i == j)
            return 0;

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

        int ans = INT_MAX;
        string s1 = "";
        for (int k = i; k <= j; k++) {
            s1 += s[k];
            if (is_palindrome(s1)) {
                if (k + 1 <= j)
                    ans = min(ans, 1 + solve(s, k + 1, j));
                else
                    ans = 0;
            }
        }

        return dp[i][j] = ans;
    }

    int minCut(string s) {
        memset(dp, -1, sizeof(dp));
        return solve(s, 0, (int)s.size() - 1);
    }
};
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