<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 DisjointSetsError(Exception):
	pass

class DisjointSetsNode:
	def __init__(self, x):
		self.value = x
		self.parent = self
		self.size = 1

	@staticmethod
	def find(x, splitting=False):
		if splitting:
			while x.parent is not x:
				x.parent, x = x.parent.parent, x.parent
			return x
		else:
			rep = x
			while rep.parent is not rep:
				rep = rep.parent
			while x is not rep:
				x.parent, x = rep, x.parent
			return rep

	@staticmethod
	def union(x, y):
		x = DisjointSetsNode.find(x)
		y = DisjointSetsNode.find(y)
		if x is y:
			return False
		if x.size < y.size:
			x, y = y, x
		y.parent = x
		x.size += y.size
		del y.size
		return True

class DisjointSets:
	def __init__(self):
		self.nodes = {}
		self.count = 0

	def make_set(self, x):
		if x in self.nodes:
			raise DisjointSetsError('Cannot add item that is already present')
		self.nodes[x] = DisjointSetsNode(x)
		self.count += 1

	def __contains__(self, x):
		return x in self.nodes

	def find(self, x, splitting=False):
		if x not in self.nodes:
			raise DisjointSetsError('Item must be present to perform find')
		return DisjointSetsNode.find(self.nodes[x], splitting).value

	def union(self, x, y):
		if x not in self.nodes or y not in self.nodes:
			raise DisjointSetsError('Both items must be present to perform union')
		merged = DisjointSetsNode.union(self.nodes[x], self.nodes[y])
		if merged:
			self.count -= 1
		return merged

	def __len__(self):
		return self.count

class Solution:
    def containsCycle(self, grid: List[List[str]]) -> bool:
        m,n = len(grid),len(grid[0])
        dj = DisjointSets()
        for i in range(m):
            for j in range(n):
                dj.make_set((i,j))
                if i:
                    if grid[i][j] == grid[i-1][j]:
                        dj.union((i,j),(i-1,j))
                if j:
                    if grid[i][j] == grid[i][j-1]:
                        if not dj.union((i,j),(i,j-1)):
                            return True
        return False
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