<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>

#include <bits/stdc++.h>
#define ss second
#define ff first
#define pii pair<int,int>
#define pb push_back
#define ll long long
#define vi vector<int>
#define vvi vector<vi>
#define vll vector<ll>	
#define pll pair<ll,ll>
#define vpll vector<pll>
#define ld long double
#define um unordered_map
#define all(x) x.begin(),x.end()

#define inf 100000000000LL

class Solution {
public:
    int minCost(vector<vector<int>>& grid) {
        ll m = grid.size();
        ll n = grid[0].size();
        vector<pll> dir = {
            {0, 1},
            {0, -1},
            {1, 0},
            {-1, 0}}; // dir[0],dir[1],...so on for values 1,2,3..

        vector<vll> d(m, vll(n, inf));
        d[0][0]=0;

        priority_queue<vll, vector<vll>, greater<vll>> pq;
        pq.push({0, 0, 0}); // top-left : cost,r,c

        while (!pq.empty()) {
            auto v = pq.top(); pq.pop();
            ll cost = v[0], r = v[1], c = v[2];
            
            if (d[r][c] != cost)
                continue;
    
            for (ll i = 0; i < 4; i++) {
                ll nr = r + dir[i].ff;
                ll nc = c + dir[i].ss;
                if (nr < 0 || nc < 0 || nr >= m || nc >= n)
                    continue;
                ll ncost = cost;
                if (i != (grid[r][c] - 1))
                    ncost += 1;

                if (ncost < d[nr][nc]) {
                    d[nr][nc] = ncost;
                    pq.push({ncost, nr, nc});
                }
            }
        }
        return d[m - 1][n - 1];
    }
};
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