<p>You are asked to cut off all the trees in a forest for a golf event. The forest is represented as an <code>m x n</code> matrix. In this matrix:</p>

<ul>
	<li><code>0</code> means the cell cannot be walked through.</li>
	<li><code>1</code> represents an empty cell that can be walked through.</li>
	<li>A number greater than <code>1</code> represents a tree in a cell that can be walked through, and this number is the tree&#39;s height.</li>
</ul>

<p>In one step, you can walk in any of the four directions: north, east, south, and west. If you are standing in a cell with a tree, you can choose whether to cut it off.</p>

<p>You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes <code>1</code> (an empty cell).</p>

<p>Starting from the point <code>(0, 0)</code>, return <em>the minimum steps you need to walk to cut off all the trees</em>. If you cannot cut off all the trees, return <code>-1</code>.</p>

<p><strong>Note:</strong> The input is generated such that no two trees have the same height, and there is at least one tree needs to be cut off.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,4],[7,6,5]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Following the path above allows you to cut off the trees from shortest to tallest in 6 steps.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/trees2.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> forest = [[1,2,3],[0,0,0],[7,6,5]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The trees in the bottom row cannot be accessed as the middle row is blocked.
</pre>

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

<pre>
<strong>Input:</strong> forest = [[2,3,4],[0,0,5],[8,7,6]]
<strong>Output:</strong> 6
<b>Explanation:</b> You can follow the same path as Example 1 to cut off all the trees.
Note that you can cut off the first tree at (0, 0) before making any steps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == forest.length</code></li>
	<li><code>n == forest[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>0 &lt;= forest[i][j] &lt;= 10<sup>9</sup></code></li>
	<li>Heights of all trees are <strong>distinct</strong>.</li>
</ul>

#define mypair pair<int, pair<int, int>>
class Solution {
public:
    int cutOffTree(vector<vector<int>>& forest) {
        int n = forest.size(), m = forest[0].size();
        vector<mypair> trees;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                if(forest[i][j] > 1) 
                    trees.push_back({forest[i][j], {i, j}});
        sort(trees.begin(), trees.end());
        /* 
        created the order to cut the tree, 
        so will find the distance to traver each of them one by one.
        */

        int steps = 0;
        int sx=0, sy=0;
        for(int i=0; i<trees.size(); i++){
            int ex = trees[i].second.first;
            int ey = trees[i].second.second;
            int req = minSteps(forest, sx, sy, ex, ey);
            if(req < 0) return -1;
            steps += req;
            sx = ex; 
            sy = ey;
        }
        return steps;
    }

    // find the minimum steps to move from {sx, sy} to {ex, ey}
    int minSteps(vector<vector<int>>& forest, int sx, int sy, int ex, int ey){
        if(sx==ex && sy==ey)return 0;
        int n = forest.size(), m = forest[0].size();
        vector<vector<bool>> vis(n, vector<bool>(m, false));

        vector<int> dx = {-1, 0, 1, 0};
        vector<int> dy = {0, 1, 0, -1};

        queue<pair<int, pair<int, int>>> q;
        q.push({0, {sx, sy}});
        vis[sx][sy]=true;
        while(!q.empty()){
            mypair node = q.front();
            q.pop();
            sx = node.second.first; sy = node.second.second;
            for(int j=0; j<4; j++){
                int nx = sx + dx[j];
                int ny = sy + dy[j];
                if(checkValid(forest, nx, ny) ){
                    if(!vis[nx][ny]){
                        vis[nx][ny] = true;
                        if(nx==ex && ny==ey){
                            forest[nx][ny]=1;
                            return node.first+1;
                        }
                        q.push({node.first+1, {nx, ny}});
                    }
                }
            }
        }
        return -1;
    }

    bool checkValid(vector<vector<int>>& forest, int x, int y){
        int n = forest.size(), m = forest[0].size();
        if(x>=0 && x<n && y>=0 && y<m && forest[x][y] >= 1) return true;
        return false;
    }
};
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