<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, grid: list[list[str]]) -> bool:
        self.grid = grid
        nrows = len(grid)
        ncols = len(grid[0])
        # First, construct the graph.  The nodes are represented
        # by (row,col) 2-tuples, the edges by a "neighbors" dictionary
        # mapping nodes to list of 4-adjacent nodes.
        self.neighbors = defaultdict(list)
        for irow in range(nrows):
            for icol in range(ncols):
                color = grid[irow][icol]
                s_irow = irow + 1
                if s_irow < nrows and color == grid[s_irow][icol]:
                    self.neighbors[(s_irow, icol)].append((irow, icol))
                    self.neighbors[(irow, icol)].append((s_irow, icol))
                e_icol = icol + 1
                if e_icol < ncols and color == grid[irow][e_icol]:
                    self.neighbors[(irow, e_icol)].append((irow, icol))
                    self.neighbors[(irow, icol)].append((irow, e_icol))
        # Second, scan through the grid, keeping track of which nodes we've
        # visited, and once we find an unvisited node, we traverse the
        # graph until we come back to a visited node, indicating that we've
        # completed a cycle.
        self.visited = [[0 for col in range(ncols)] for row in range(nrows)]
        self.trace_index = 0
        for irow in range(nrows):
            for icol in range(ncols):
                if self.visited[irow][icol] == 0:
                    self.visited[irow][icol] = 1
                    if self.traceCycle(irow, icol):
                        print("found a cycle starting at", irow, icol)
                        return True
        return False

    # Try to trace a cycle of like-colored cells in "grid" and return True iff found:
    def traceCycle(self, irow, icol):
        self.trace_index += 1
        prevRowCol = None
        result = self.traceCycleRec((irow, icol), prevRowCol, self.trace_index)
        return result

    def traceCycleRec(self, rowCol, prevRowCol, trace_index):
        # print("self.grid\n", np.array(self.grid))
        # print("self.visited\n", np.array(self.visited))
        result = False
        for neighbor in self.neighbors[rowCol]:
            if prevRowCol == neighbor:
                continue
            neighbor_visited = self.visited[neighbor[0]][neighbor[1]]
            if neighbor_visited:
                if neighbor_visited == trace_index:
                    result = True
                break
            else:
                self.visited[neighbor[0]][neighbor[1]] = trace_index
                what = self.traceCycleRec(neighbor, rowCol, trace_index)
                # self.visited[ neighbor[ 0 ]][ neighbor[ 1 ] ] = 0
                result = what #  True
        return result
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