<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 bfs(int node, vector<vector<int>>count, int curr = 0) {
        int n = count.size();
        if (node == n)  return curr;
        int k = INT_MAX;
        for (auto x: count[node]) {
            k = min(k, bfs(x+1, count, curr+1));
        }
        k = min(bfs(node+1, count, curr+1), k);
        return k;
    }
public:
    int minCut(string s) {
        int n = s.length();
        vector<vector<bool>>dp(n, vector<bool>(n, false));
        for (int i=0; i<n; i++) dp[i][i] = true;
        vector<vector<int>>count(n);
        for (int i=n-1; i>=0; i--) {
            for (int j=i; j<n; j++) {
                if (s[i] == s[j] && ((j - i + 1) <= 3 || dp[i+1][j-1])) {
                    dp[i][j] = true;
                    count[i].push_back(j);
                }
            }
        }
        queue<int>q;
        q.push(0);
        vector<bool>vis(n, false);
        int cuts = 0;
        vis[0] = true;
        while (!q.empty()) {
            int sz = q.size();
            while (sz--) {
                int t = q.front();
                q.pop();
                for (auto x: count[t]) {
                    if (vis[x+1])  continue;
                    vis[x+1] = true;
                    q.push(x+1);
                    if ((x+1) == n) return cuts;
                }
            }
            cuts+=1;
        }
        return cuts;
    }
};
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