<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 BFS(vector<vector<char>>& grid, vector<vector<bool>>& visited, int row, int col, char ch) {
        visited[row][col] = true;
        queue<pair<int,int>> q;
        q.push({row,col});

        int n = grid.size();
        int m = grid[0].size();

        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

        map<pair<int,int>, pair<int,int>> parent;


        while(!q.empty()) {
            int currentRow = q.front().first;
            int currentCol = q.front().second;
            q.pop();
            for(auto dir : directions) {
                int newRow = currentRow + dir.first;
                int newCol = currentCol + dir.second;

                if(newRow >= 0 && newRow < n && newCol >= 0 && newCol < m){
                    if(grid[newRow][newCol] == ch && visited[newRow][newCol] == false) {
                        q.push({newRow, newCol});
                        visited[newRow][newCol] = true;
                        parent[{newRow,newCol}] = {currentRow,currentCol};
                    }
                    else if(grid[newRow][newCol] == ch && visited[newRow][newCol] && 
                   (parent[{currentRow,currentCol}].first != newRow || parent[{currentRow,currentCol}].second != newCol)) {
                        return true;
                    }
                }

            }
        }

        return false;
    }
    bool containsCycle(vector<vector<char>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        if(n == 1 || m == 1) return false;
        vector<vector<bool>> visited (n, vector<bool>(m,false));

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(visited[i][j] == false) {
                    if(BFS(grid,visited,i,j,grid[i][j])) {
                        return true;
                    }
                }
            }
        }
        return false;        
    }
};
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