<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 Solution {
public:
    bool isConnectedToParent(vector<vector<int>>& adj,int c,int x)
    {
        for(auto y:adj[x]){if(y==c)return 1;}
        return 0;
    }
    void display_adj_list(vector<vector<int>>& adj)
    {
        int nn=adj.size();
        for(int i=0;i<nn;i++)
        {
            cout<<i<<"->";
            for(auto x:adj[i])
            cout<<x<<" ";
            cout<<endl;
        }
    }
void form_adj_list(vector<vector<int>>& grid,vector<vector<int>>& adj,vector<vector<int>>& node)
    {
        int n=grid.size(),m=grid[0].size();
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                int x=grid[i][j],nod=node[i][j];
                if(x==1)
                {
                    if(j>0)
                    adj[nod].push_back(node[i][j-1]);
                    if(j<m-1)
                    adj[nod].push_back(node[i][j+1]);
                }
                else if(x==2)
                {
                    if(i>0)
                    adj[nod].push_back(node[i-1][j]);
                    if(i<n-1)
                    adj[nod].push_back(node[i+1][j]);
                }
                else if(x==3)
                { 
                    if(j>0)
                    adj[nod].push_back(node[i][j-1]);
                    if(i<n-1)
                    adj[nod].push_back(node[i+1][j]);
                }
                else if(x==4)
                {
                    if(j<m-1)
                    adj[nod].push_back(node[i][j+1]);
                    if(i<n-1)
                    adj[nod].push_back(node[i+1][j]);
                }
                else if(x==5)
                {
                    if(j>0)
                    adj[nod].push_back(node[i][j-1]);
                    if(i>0)
                    adj[nod].push_back(node[i-1][j]);
                }
                else
                {
                    if(j<m-1)
                    adj[nod].push_back(node[i][j+1]);
                    if(i>0)
                    adj[nod].push_back(node[i-1][j]);
                }
            }
        }
    }

    bool canReachDestination(vector<vector<int>>& adj)
    {
        int nn=adj.size();
        queue<int> q;
        vector<int> vis(nn,0);
        q.push(0);
        vis[0]=1;
        while(!q.empty())
        {
            int curr=q.front();
            q.pop();

            for(auto x:adj[curr])
            {
                if(!vis[x] && isConnectedToParent(adj,curr,x))
                {
                    vis[curr]=1;
                    q.push(x);
                    if(x==nn-1)
                    return 1;
                }
            }
        }
        return 0;
    }

    bool hasValidPath(vector<vector<int>>& grid) {
        int n=grid.size(),m=grid[0].size(),nn=n*m,k=0;
        if(nn==1)return 1;
        vector<vector<int>> node(n,vector<int> (m)),adj(nn);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                node[i][j]=k;
                k++;
            }
        }
        //making adjejency list
        form_adj_list(grid,adj,node);
        //display adj list
        //display_adj_list(adj);
        //is valid path to reach destination using BFS
        return canReachDestination(adj);
    }
};
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