<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>

class Solution {
public:
    bool containsCycle(vector<vector<char>>& grid) {
        queue<pair<int,int>> q;
        int count =0;
        pair<int,int> previous[grid.size()][grid[0].size()];
        int  visited[501][501] = {0};
        for(int i = 0 ; i  < grid.size();i++){
            for(int j = 0 ; j < grid[0].size(); j++){ 
                if(!visited[i][j]){
                    q.push(make_pair(i,j));
                    visited[i][j]=1;
                    previous[i][j] = make_pair(-1, -1);
                    while(!q.empty()){
                        if(q.front().first-1>=0&&grid[q.front().first-1][q.front().second] == grid[q.front().first][q.front().second]){

                            if(!visited[q.front().first-1][q.front().second]){
                                q.push(make_pair(q.front().first-1,q.front().second));
                                visited[q.front().first-1][q.front().second] =1;
                                previous[q.front().first-1][q.front().second] = make_pair(q.front().first,q.front().second);
                            }
                            else if(visited[q.front().first-1][q.front().second]&&(previous[q.front().first][q.front().second].first!=q.front().first-1||previous[q.front().first][q.front().second].second!=q.front().second)){
                               
                                return true;
                            }
                        }
                        if(q.front().second-1>=0&&grid[q.front().first][q.front().second-1] == grid[q.front().first][q.front().second]){
                            if(!visited[q.front().first][q.front().second-1]){
                                q.push(make_pair(q.front().first,q.front().second-1));
                                visited[q.front().first][q.front().second-1] =1;
                                previous[q.front().first][q.front().second-1] = make_pair(q.front().first,q.front().second);
                                
                            }
                            else if(visited[q.front().first][q.front().second-1]&&(previous[q.front().first][q.front().second].first!=q.front().first||previous[q.front().first][q.front().second].second!=q.front().second-1)){
                                
                                return true;
                            }
                        }
                        if(q.front().first+1<grid.size()&&grid[q.front().first+1][q.front().second] == grid[q.front().first][q.front().second]){
                            if(!visited[q.front().first+1][q.front().second]){
                                q.push(make_pair(q.front().first+1,q.front().second));
                                visited[q.front().first+1][q.front().second] =1;
                                 previous[q.front().first+1][q.front().second] = make_pair(q.front().first,q.front().second);
                            }
                            else if(visited[q.front().first+1][q.front().second]&&(previous[q.front().first][q.front().second].first!=q.front().first+1||previous[q.front().first][q.front().second].second!=q.front().second)){
                                return true;
                            }
                        }
                        if(q.front().second+1<grid[0].size()&&grid[q.front().first][q.front().second+1] == grid[q.front().first][q.front().second]){
                            if(!visited[q.front().first][q.front().second+1]){
                                q.push(make_pair(q.front().first,q.front().second+1));
                                visited[q.front().first][q.front().second+1] =1;
                                 previous[q.front().first][q.front().second+1] = make_pair(q.front().first,q.front().second);
                                
                            }
                            else if(visited[q.front().first][q.front().second+1]&&(previous[q.front().first][q.front().second].first!=q.front().first||previous[q.front().first][q.front().second].second!=q.front().second+1)){
                                return true;
                            }
                        }
                        //cout<<q.front().first<<","<<q.front().second<<" ";
                        q.pop();

                    }
                    cout<<endl;
                }
            }
        }
    return false;
    }
    
};
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