<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>

class Solution {
public:
    int bfs(vector<vector<int>>& forest,vector<vector<bool>>& visited, int destx, int desty, int currx, int curry)
    {
        int currstep = 0;
        int dest = forest[destx][desty];
        queue<pair<int, int>> q;
        q.push({currx, curry});
        while(q.size()>0)
        {
            int currsize = q.size();
            for(int i = 0; i<currsize;i++)
            {
                pair<int, int> curr = q.front();
                q.pop();
                if(curr.first == destx && curr.second == desty) return currstep;
                visited[curr.first][curr.second] = true;
                if(curr.first-1 >=0)
                {
                    if(forest[curr.first-1][curr.second] == dest) return currstep+1;
                    if(visited[curr.first-1][curr.second] == false && forest[curr.first-1][curr.second] > 0) 
                        {
                            visited[curr.first-1][curr.second] = true;
                            q.push({curr.first-1, curr.second});
                        }
                }
                if(curr.first+1 < forest.size())
                {
                    if(forest[curr.first+1][curr.second] == dest) return currstep+1;
                    if(visited[curr.first+1][curr.second] == false && forest[curr.first+1][curr.second] > 0) 
                    {
                        q.push({curr.first+1, curr.second});
                        visited[curr.first+1][curr.second] = true;
                    }
                        
                }
                if(curr.second-1 >=0)
                {
                    if(forest[curr.first][curr.second-1] == dest) return currstep+1;
                    if(visited[curr.first][curr.second-1] == false && forest[curr.first][curr.second-1] > 0) 
                        
                        {
                            visited[curr.first][curr.second-1] = true;

                            q.push({curr.first, curr.second-1});
                        }
                }
                if(curr.second+1 < forest[0].size())
                {
                    if(forest[curr.first][curr.second+1] == dest) return currstep+1;
                    if(visited[curr.first][curr.second+1] == false && forest[curr.first][curr.second+1] > 0) 
                    {
                        q.push({curr.first, curr.second+1});
                        visited[curr.first][curr.second+1] = true;
                    }
                }
            }
            currstep++;
        }
        return -1;
    } 

    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size();
        int n = forest[0].size();
        vector<int> trees;
        map<int, pair<int, int>> mp;
        int steps = 0;
        for(int i = 0; i< m; i++)
        {
            for(int j = 0; j< n; j++)
            {
                if(forest[i][j] > 1) 
                {
                    trees.push_back(forest[i][j]);
                    mp[forest[i][j]] = {i, j};
                }
            }
        }
        sort(trees.begin(), trees.end());
        int currx = 0;
        int curry = 0;
        for(int i = 0; i < trees.size(); i++)
        {
            int destx = mp[trees[i]].first;
            int desty = mp[trees[i]].second;
            vector<vector<bool>> visited(m, vector<bool>(n, false));
            int curr = bfs(forest, visited, destx, desty, currx, curry);
            if(curr == -1) return -1;
            else steps+=curr;
            currx = destx;
            curry = desty;
        }

        return steps;
    }
};
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