<p>Given an <code>m x n</code> grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of <code>grid[i][j]</code> can be:</p>

<ul>
	<li><code>1</code> which means go to the cell to the right. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j + 1]</code>)</li>
	<li><code>2</code> which means go to the cell to the left. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j - 1]</code>)</li>
	<li><code>3</code> which means go to the lower cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i + 1][j]</code>)</li>
	<li><code>4</code> which means go to the upper cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i - 1][j]</code>)</li>
</ul>

<p>Notice that there could be some signs on the cells of the grid that point outside the grid.</p>

<p>You will initially start at 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> following the signs on the grid. The valid path does not have to be the shortest.</p>

<p>You can modify the sign on a cell with <code>cost = 1</code>. You can modify the sign on a cell <strong>one time only</strong>.</p>

<p>Return <em>the minimum cost to make the grid have at least one valid path</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid1.png" style="width: 400px; height: 390px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --&gt; (0, 1) --&gt; (0, 2) --&gt; (0, 3) change the arrow to down with cost = 1 --&gt; (1, 3) --&gt; (1, 2) --&gt; (1, 1) --&gt; (1, 0) change the arrow to down with cost = 1 --&gt; (2, 0) --&gt; (2, 1) --&gt; (2, 2) --&gt; (2, 3) change the arrow to down with cost = 1 --&gt; (3, 3)
The total cost = 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid2.png" style="width: 350px; height: 341px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,3],[3,2,2],[1,1,4]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> You can follow the path from (0, 0) to (2, 2).
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid3.png" style="width: 200px; height: 192px;" />
<pre>
<strong>Input:</strong> grid = [[1,2],[4,3]]
<strong>Output:</strong> 1
</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;= 100</code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 4</code></li>
</ul>

// https://leetcode.com/problems/minimum-cost-to-make-at-least-one-valid-path-in-a-grid/?envType=problem-list-v2&envId=shortest-path
typedef long long ll;
class Solution {
public:
    vector<int> rows = {1, -1, 0, 0};
    vector<int> cols = {0, 0, 1, -1};
    int minCost(vector<vector<int>>& grid) {
        int m=grid.size(),n=grid[0].size();
        vector<vector<ll> > dp(m,vector<ll> (n,LLONG_MAX)); //dp[i][j] : min cost to reach i,j

        queue<pair<int,int> > q;
        q.emplace(0,0);
        dp[0][0]=0;
        unordered_map<int,pair<int,int> > arrowDirs;
        arrowDirs[1]={0,1}; arrowDirs[2]={0,-1}; arrowDirs[3]={1,0}; arrowDirs[4]={-1,0};
        while(!q.empty())
        {
            int x=q.front().first;
            int y=q.front().second;
            q.pop();
            for(int i=0;i<4;i++)
            {
                int nx=x+rows[i];
                int ny=y+cols[i];

                if(nx<0 || nx>=m || ny<0 || ny>=n)  
                    continue;

                ll cost = (arrowDirs[grid[x][y]] == make_pair(rows[i], cols[i])) ? 0 : 1;

                if (dp[x][y] + cost < dp[nx][ny]) {
                    dp[nx][ny] = dp[x][y] + cost;
                    q.emplace(nx, ny);
                }
                
                if(cost + dp[x][y] < dp[nx][ny])
                {
                    dp[nx][ny]=cost+dp[x][y];
                    q.emplace(nx,ny);
                }
            }
        }

        return dp[m-1][n-1];
    }
};

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