<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:
    def containsCycle(self, arr):
        """
        :type grid: List[List[str]]
        :rtype: bool
        """
        matrix_visited = []
        for row in range(0, len(arr)):
            temp = []
            for col in range(0, len(arr[0])):
                temp.append(False)
            matrix_visited.append(temp)
                
        for row in range(0, len(arr)):
            for col in range(0, len(arr[0])):
                if matrix_visited[row][col] == True: continue
                visited = set()
                if self.dfs(arr, row, col,-1, -1, visited, arr[row][col], matrix_visited):
                    return True
        return False
        
    def dfs(self, arr, row, col,prv_r, prv_c, visited, current, matrix_visited):
        if (row, col) in visited: return True
        visited.add((row, col))
        matrix_visited[row][col] = True

        for (r, c) in [(-1, 0), (1, 0)]:
            r,c = row + r, col + c
            rowInbound = r >= 0 and r < len(arr)
            colInbound = c >= 0 and c < len(arr[0])
            if not rowInbound or not colInbound: continue
            if r == prv_r: continue
            if arr[r][c] == current:
                res = self.dfs(arr, r, c, row, col, visited, arr[r][c], matrix_visited)
                if res:
                    return True
                visited.remove((r, c))

        for (r, c) in [(0, -1), (0, 1)]:
            r,c = row + r, col + c
            rowInbound = r >= 0 and r < len(arr)
            colInbound = c >= 0 and c < len(arr[0])
            if not rowInbound or not colInbound: continue
            if c == prv_c: continue
            if arr[r][c] == current:
                res = self.dfs(arr, r, c, row, col, visited, arr[r][c], matrix_visited)
                if res:
                    return True
                visited.remove((r, c))
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