<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 isPali(string& s, int i, int j) {
        while (i < j) {
            if (s[i] != s[j]) return false;
            i++;
            j--;
        }
        return true;
    }

    int minCut(string s) {

        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> pq;
        pq.push({0, 0});
        vector<int> vis(s.size(), 0);

        while(!pq.empty()) {
            auto pr = pq.top();
            pq.pop();

            int idx = pr.second;
            int cuts = pr.first;

            if (idx == s.size()) return cuts-1;
            if (vis[idx]) continue;
            vis[idx] = 1;

            for (int i = idx; i < s.length(); i++) {
                if (isPali(s, idx, i)) {
                    pq.push({cuts+1, i+1});
                }
            }
        }
        return -1;
    }
};
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