<p>You are given an <code>m x n</code> <code>grid</code>. Each cell of <code>grid</code> represents a street. The street of <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means a street connecting the left cell and the right cell.</li>
	<li><code>2</code> which means a street connecting the upper cell and the lower cell.</li>
	<li><code>3</code> which means a street connecting the left cell and the lower cell.</li>
	<li><code>4</code> which means a street connecting the right cell and the lower cell.</li>
	<li><code>5</code> which means a street connecting the left cell and the upper cell.</li>
	<li><code>6</code> which means a street connecting the right cell and the upper cell.</li>
</ul>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/main.png" style="width: 450px; height: 708px;" />
<p>You will initially start at the street of the upper-left cell <code>(0, 0)</code>. A valid path in the grid is a path that starts from the upper left cell <code>(0, 0)</code> and ends at the bottom-right cell <code>(m - 1, n - 1)</code>. <strong>The path should only follow the streets</strong>.</p>

<p><strong>Notice</strong> that you are <strong>not allowed</strong> to change any street.</p>

<p>Return <code>true</code><em> if there is a valid path in the grid or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/e1.png" style="width: 455px; height: 311px;" />
<pre>
<strong>Input:</strong> grid = [[2,4,3],[6,5,2]]
<strong>Output:</strong> true
<strong>Explanation:</strong> As shown you can start at cell (0, 0) and visit all the cells of the grid to reach (m - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/05/e2.png" style="width: 455px; height: 293px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,1],[1,2,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> As shown you the street at cell (0, 0) is not connected with any street of any other cell and you will get stuck at cell (0, 0)
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,2]]
<strong>Output:</strong> false
<strong>Explanation:</strong> You will get stuck at cell (0, 1) and you cannot reach cell (0, 2).
</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;= 300</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 6</code></li>
</ul>


class DisjointSet{
public:
    vector<int>par,size;
    DisjointSet(int n){
        par.resize(n+1);
        size.resize(n+1,1);
        for(int i=0; i<=n; i++){
            par[i] = i;
        }
    }
    int findpar(int u){
        if(u == par[u]) return u;
        return par[u] = findpar(par[u]);
    }

    void unionBySize(int u, int v){
        u = findpar(u);
        v = findpar(v);

        if(u == v) return;

        if(size[u] >= size[v]){
            par[v] = u;
            size[u] += size[v];
        }
        else{
            par[u] = v;
            size[v] += size[u];
        }
    }
    
};
class Solution {
public:
    bool hasValidPath(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        DisjointSet ds(n*m);
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                int p = i * m + j;
                int p1 = (i-1)*m + j;  // 1
                int p2 = (i+1)*m + j;  // 1
                int p3 = i*m + (j+1);  // 2
                int p4 = i*m + (j-1);  // 2

                if(grid[i][j] == 1){
                    if(j+1 < m && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) ){
                        ds.unionBySize(p,p3);
                    }
                    if(j-1 >= 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) ){
                        ds.unionBySize(p,p4);
                    }
                }
                else if(grid[i][j] == 2){
                    if(i+1 < n && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) ){
                        ds.unionBySize(p,p2);
                    }

                    if(i-1>=0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4)){
                        ds.unionBySize(p,p1);
                    }
                }
                else if(grid[i][j] == 3){
                    if(j-1 >= 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) ){
                        ds.unionBySize(p,p4);
                    }
                    if(i+1 < n && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) ){
                        ds.unionBySize(p,p2);
                    }
                }
                else if(grid[i][j] == 4){
                    if(j+1 < m && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) ){
                        ds.unionBySize(p,p3);
                    }
                    if(i+1 < n && (grid[i+1][j] == 2 || grid[i+1][j] == 5 || grid[i+1][j] == 6) ){
                        ds.unionBySize(p,p2);
                    }
                }
                else if(grid[i][j] == 5){
                    if(i-1>=0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4)){
                        ds.unionBySize(p,p1);
                    }
                    if(j-1 >= 0 && (grid[i][j-1] == 1 || grid[i][j-1] == 4 || grid[i][j-1] == 6) ){
                        ds.unionBySize(p,p4);
                    }
                }
                else if(grid[i][j] == 6){
                    if(i-1>=0 && (grid[i-1][j] == 2 || grid[i-1][j] == 3 || grid[i-1][j] == 4)){
                        ds.unionBySize(p,p1);
                    }
                    if(j+1 < m && (grid[i][j+1] == 1 || grid[i][j+1] == 3 || grid[i][j+1] == 5) ){
                        ds.unionBySize(p,p3);
                    }
                }
            }
        }
        return ds.findpar(0) == ds.findpar(n*m-1);
    }
};

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