<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:
    int n, m;
    bool ans = false;
    vector<vector<bool>> visited;
    vector<vector<char>> a;
    vector<pair<int, int> > b;

    void dfs(int i, int j, int value, int pre_i, int pre_j) {
        cout << i << " " << j << " " << value << visited[i][j] << "\n";
        if (visited[i][j] == true) {
            cout << "cycle";
            ans = true;
            return;    
        }
        
        visited[i][j] = true;
        int dx[] = {0, 0, -1, 1};
        int dy[] = {-1, 1, 0, 0};
        for (int k = 0; k <= 3; k++) {
            int newx = i + dx[k];
            int newy = j + dy[k];
            // cout << "new: " << newx << " " << newy << "\n";
            if (newx < 0 || newx >= n) continue;
            // cout << "1";
            if (newy < 0 || newy >= m) continue;
            // cout << "2";
            if (a[newx][newy] != value) continue;
            // cout << "3";
            if (newx != pre_i || newy != pre_j) dfs(newx, newy, value, i, j);
        }
    }

    bool containsCycle(vector<vector<char>>& grid) {
        n = grid.size();
        m = grid[0].size();
        visited.resize(n);
        a = grid;
        for (auto& tmp: visited) {
            tmp.resize(m);
        }
        // dfs(0, 0, a[0][0], -1, -1);
        ans = false;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (visited[i][j] == true) continue;
                dfs(i, j, a[i][j], -1, -1);
                if (ans == true) return true;
            }
        }
        return false;
    }
};
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