<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>

#include <vector>
#include <tuple>
#include <queue>

using namespace std;

int moveX[] = { -1, 1, 0, 0 };
int moveY[] = { 0, 0, -1, 1 };

struct UnionFind {
    vector<int> parent, rank;

    UnionFind(int n) : parent(n), rank(n, 1) {
        for (int i = 0; i < n; ++i) {
            parent[i] = i;
        }
    }

    int find(int u) {
        if (parent[u] != u) {
            parent[u] = find(parent[u]);
        }
        return parent[u];
    }

    bool unionSet(int u, int v) {
        int rootU = find(u);
        int rootV = find(v);
        if (rootU == rootV) {
            return false;
        }
        if (rank[rootU] > rank[rootV]) {
            parent[rootV] = rootU;
        } else if (rank[rootU] < rank[rootV]) {
            parent[rootU] = rootV;
        } else {
            parent[rootV] = rootU;
            rank[rootU]++;
        }
        return true;
    }
};

class Solution {
public:
    int N, M;
    using Visited = vector<vector<bool>>;
    Visited visited;

    bool dfs(int startX, int startY, int px, int py, UnionFind &unionFind, vector<vector<char>>& grid) {
        visited[startX][startY] = true;
        int u = startX * M + startY;

        for (int i = 0; i < 4; ++i) {
            int xx = startX + moveX[i];
            int yy = startY + moveY[i];

            if (xx < 0 || xx >= N || yy < 0 || yy >= M) {
                continue;
            }

            if (grid[xx][yy] != grid[startX][startY]) {
                continue;
            }

            int v = xx * M + yy;

            if (!visited[xx][yy]) {
                if (!unionFind.unionSet(u, v)) {
                    return true; // Phát hiện chu trình
                }
                if (dfs(xx, yy, startX, startY, unionFind, grid)) {
                    return true;
                }
            } else {
                
                int p = px * M + py;
                

                // if(!= unionFind.find(p)) {
                //     return true;
                // }

                if (v != px * M + py ) {
                    return true; // Phát hiện chu trình
                }
            }
        }

        return false;
    }

    bool containsCycle(vector<vector<char>>& grid) {
        N = grid.size();
        M = grid[0].size();

        visited = Visited(N, vector<bool>(M, false));
        UnionFind unionFind(N * M);

        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < M; ++j) {
                if (!visited[i][j]) {
                    if (dfs(i, j, -1, -1, unionFind, grid)) {
                        return true;
                    }
                }
            }
        }

        return false;
    }
};

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