<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 isPalindrome(string& s, int start, int end)
    {
        while (start <= end)
        {
            if (s[start++] != s[end--]) return false;
        }
        return true;
    }

    int minCut(string s)
    {
        int n = s.size();
        queue<pair<int, int>> qu;
        vector<bool> visited(n, false);
        qu.emplace(0, 0);

        while (!qu.empty())
        {
            auto [start, cuts] = qu.front(); qu.pop();

            for (int end = n - 1; end >= start; end--)
            {
                if (!visited[end] && isPalindrome(s, start, end)) // It can be possbile that start and end both are equal
                {                                                 // so do set visited[start] = true, after for loop
                    if (end == n - 1) return cuts;
                    qu.emplace(end + 1, cuts + 1);
                    //cout << end << endl;
                }
            }
            visited[start] = true;
        }
        return 0;
    }
};

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