<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:
    bool containsCycle(vector<vector<char>>& grid) {
    int M = grid.size();
    int N = grid[0].size();
    vector<bool> unvisited(N*M, true);
    for(int cur_it = 0; cur_it < unvisited.size(); cur_it++)
    {
        if(!unvisited[cur_it])
            continue;
        int cur_x = (cur_it) / N;
        int cur_y = (cur_it) % N;
        char cur_char = grid[cur_x][cur_y];
        unvisited[cur_it] = false;
        queue<int> toVisit;
        if (cur_x > 0 && grid[cur_x - 1][cur_y] == cur_char)
            toVisit.push(cur_it - N);
        if (cur_x < M - 1 && grid[cur_x + 1][cur_y] == cur_char)
            toVisit.push(cur_it + N);
        if (cur_y > 0 && grid[cur_x][cur_y - 1] == cur_char)
            toVisit.push(cur_it - 1);
        if (cur_y < N - 1 && grid[cur_x][cur_y + 1] == cur_char)
            toVisit.push(cur_it + 1);
        while (!toVisit.empty())
        {
            int cur_ind = toVisit.front();
            toVisit.pop();
            cur_x = cur_ind / N;
            cur_y = cur_ind % N;
            if (!unvisited[cur_ind] && grid[cur_x][cur_y] == cur_char)
                return true;
            unvisited[cur_ind] = false;
            if (cur_x > 0 && grid[cur_x - 1][cur_y] == cur_char && unvisited[cur_ind - N])
                toVisit.push(cur_ind - N);
            if (cur_x < M - 1 && grid[cur_x + 1][cur_y] == cur_char && unvisited[cur_ind + N])
                toVisit.push(cur_ind + N);
            if (cur_y > 0 && grid[cur_x][cur_y - 1] == cur_char && unvisited[cur_ind - 1])
                toVisit.push(cur_ind - 1);
            if (cur_y < N - 1 && grid[cur_x][cur_y + 1] == cur_char && unvisited[cur_ind +1])
                toVisit.push(cur_ind + 1);
        };
    }
    return false;
    }
};
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