<p>Given a 2D array of characters <code>grid</code> of size <code>m x n</code>, you need to find if there exists any cycle consisting of the <strong>same value</strong> in <code>grid</code>.</p>

<p>A cycle is a path of <strong>length 4 or more</strong> in the grid that starts and ends at the same cell. From a given cell, you can move to one of the cells adjacent to it - in one of the four directions (up, down, left, or right), if it has the <strong>same value</strong> of the current cell.</p>

<p>Also, you cannot move to the cell that you visited in your last move. For example, the cycle <code>(1, 1) -&gt; (1, 2) -&gt; (1, 1)</code> is invalid because from <code>(1, 2)</code> we visited <code>(1, 1)</code> which was the last visited cell.</p>

<p>Return <code>true</code> if any cycle of the same value exists in <code>grid</code>, otherwise, return <code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/1.png" style="width: 231px; height: 152px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;a&quot;,&quot;a&quot;,&quot;a&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;,&quot;a&quot;],[&quot;a&quot;,&quot;a&quot;,&quot;a&quot;,&quot;a&quot;]]
<strong>Output:</strong> true
<strong>Explanation: </strong>There are two valid cycles shown in different colors in the image below:
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/11.png" style="width: 225px; height: 163px;" />
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/22.png" style="width: 236px; height: 154px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;c&quot;,&quot;c&quot;,&quot;c&quot;,&quot;a&quot;],[&quot;c&quot;,&quot;d&quot;,&quot;c&quot;,&quot;c&quot;],[&quot;c&quot;,&quot;c&quot;,&quot;e&quot;,&quot;c&quot;],[&quot;f&quot;,&quot;c&quot;,&quot;c&quot;,&quot;c&quot;]]
<strong>Output:</strong> true
<strong>Explanation: </strong>There is only one valid cycle highlighted in the image below:
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/2.png" style="width: 229px; height: 157px;" />
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/07/15/3.png" style="width: 183px; height: 120px;" /></strong></p>

<pre>
<strong>Input:</strong> grid = [[&quot;a&quot;,&quot;b&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;z&quot;,&quot;b&quot;],[&quot;b&quot;,&quot;b&quot;,&quot;a&quot;]]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid</code> consists only of lowercase English letters.</li>
</ul>

struct Disjoint{
   vector<int> parent;
   Disjoint(int n){
    parent.resize(n);
    for(int i{};i<n;++i)parent[i]=i;
   }
   int findParent(int x){
        if(parent[x]!=x){
            parent[x]=findParent(parent[x]);
        }
        return parent[x];
   }
   void merge(int x,int y){
       int a = findParent(x);
       int b = findParent(y);
       parent[b]=a;
        return;
   }
};
class Solution {
public:
    bool containsCycle(vector<vector<char>>& grid) {
        int dx[4]={1,0};
       int dy[4]={0,1};
       int n = grid.size();int m = grid[0].size();
       vector<vector<int>> edges;
       vector<int> temp(4,0);
       for(int i{};i<n;++i)
       for(int j{};j<m;++j){
            for(int k{};k<2;++k){
                int a = i + dx[k];
                int b = j + dy[k];
                if(0<=a && a<n && 0<=b && b<m && grid[i][j]==grid[a][b]){
                    temp = {i,j,a,b};
                    edges.push_back(temp);
                }
            }
       }
       Disjoint* disjoint = new Disjoint(n*m);
       //cout<<edges.size()<<endl;
       for(int i{};i<edges.size();++i){
            int a = disjoint->findParent(edges[i][0]*m + edges[i][1]);
            int b = disjoint->findParent(edges[i][2]*m + edges[i][3]);
            if(a==b){
                return true;
            }
            disjoint ->merge(a,b);
       }
       return false;
    }
};
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