<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 cutOffTree(vector<vector<int>>& forest) {
        map<int, vector<int>> trees;

        // find location of all trees to cut
        for(int i = 0; i < forest.size(); i++){
            for(int j = 0; j<forest[0].size(); j++){
                if(forest[i][j] > 1) trees[forest[i][j]] = {i,j};
            }
        }
        
        int total_distance = 0;
        
        vector<int> current_location{0,0};
        //find path to smallest tree.
        while(!trees.empty()){
            // ordered map begins with smallest key
            vector<int> smallest_tree_location = trees.begin()->second; //get location of smallest tree
            trees.erase(trees.begin());
            int distance = bfs(forest, current_location, smallest_tree_location);
            if(distance < 0) return -1;
            total_distance += distance;
            current_location = smallest_tree_location;
        }
        
        return total_distance;
    }
    
    
    int bfs(vector<vector<int>>& grid, vector<int> start, vector<int>end){
        
        int rows = grid.size();
        int cols = grid[0].size();
        
        vector<vector<int>> visited(rows, vector<int>(cols, 0));        
        
        int distance = 0;
        
        queue<pair<int,int>> q;
        
        q.push(make_pair(start[0],start[1]));
        
        while(!q.empty()){
            int size = q.size();
            
            // Distance increments after each level
            while(size--){
                auto loc = q.front();
                q.pop();
                
                // Access pair contents using inbuilt methods first and second
                if(loc.first == end[0] && loc.second == end[1]) return distance;
                
                // move up
                if(loc.first != 0){
                    if(grid[loc.first-1][loc.second] != 0 && visited[loc.first-1][loc.second] == 0){
                        visited[loc.first-1][loc.second] = 1;
                        q.push(make_pair(loc.first-1,loc.second));
                    }
                }

                // move down
                if(loc.first != rows-1){
                    if(grid[loc.first+1][loc.second] != 0 && visited[loc.first+1][loc.second] == 0){
                        visited[loc.first+1][loc.second] = 1;
                        q.push(make_pair(loc.first+1,loc.second));
                    }
                }

                // move left
                if(loc.second != 0){
                    if(grid[loc.first][loc.second-1] != 0 && visited[loc.first][loc.second-1] == 0){
                        visited[loc.first][loc.second-1] = 1;
                        q.push(make_pair(loc.first,loc.second-1));
                    }
                }

                // move right
                if(loc.second != cols-1){
                    if(grid[loc.first][loc.second+1] != 0 && visited[loc.first][loc.second+1] == 0){
                        visited[loc.first][loc.second+1] = 1;
                        q.push(make_pair(loc.first,loc.second+1));
                    }
                }
            }
            distance++;
        }
        
        return -1;
    }
};
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