<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 {
private:
    int dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};

public:
    bool isValid(int i, int j, vector<vector<char>>& grid) {
        return i >= 0 && j >= 0 && i < grid.size() && j < grid[0].size();
    }

    bool dfs(int i, int j, int pi, int pj, char current,
             vector<vector<char>>& grid, vector<vector<bool>>& visited,
             int length) {
        visited[i][j] = true;
        for (int k = 0; k < 4; ++k) {
            int nx = i + dx[k], ny = j + dy[k];
            // Ensure next position is valid and not the previous cell
            if (isValid(nx, ny, grid) && grid[nx][ny] == current) {
                if (!visited[nx][ny]) {
                    if (dfs(nx, ny, i, j, current, grid, visited, length + 1)) {
                        return true;
                    }
                } else if (nx != pi || ny != pj) { // didn't find a cycle
                    if (length >= 4) { // Check if the cycle length is >= 4
                        return true;
                    }
                }
            }
        }
        return false;
    }

    bool containsCycle(vector<vector<char>>& grid) {
        vector<vector<bool>> visited(grid.size(),
                                     vector<bool>(grid[0].size(), false));
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (!visited[i][j]) {
                    if (dfs(i, j, -1, -1, grid[i][j], grid, visited, 1)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
};
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