<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 stepsToCutNextTree(int startX, int startY, int n, int m, vector<vector<int>>& forest, int targX, int targY)
    {
        if((startX==targX)&&(startY==targY))
            return 0;
        
        
        queue<tuple<int, int,int>> q;
        q.push({startX, startY,0});
        
        vector<vector<int>> visited(n,vector<int>(m,0));

      

        int steps=0;
        
        while(!q.empty())
        {
            auto[x,y,dist] = q.front();
            
            q.pop();
            if((x==targX)&&(y==targY))
                return dist;
            
            if((x<0) || (x>n-1) || (y<0) || (y >m-1)||(forest[x][y]==0) || (visited[x][y]==1))
                continue;

            visited[x][y] = 1;
            
            q.push({x+1,y,dist+1});
            q.push({x-1,y,dist+1});
            q.push({x,y+1,dist+1});
            q.push({x,y-1,dist+1});

          
        }

        return -1;

    }
    int cutOffTree(vector<vector<int>>& forest) {
        vector<tuple<int,int,int>> target_trees;
        int n = forest.size();
        int m = forest[0].size();
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(forest[i][j] > 1)
                    target_trees.push_back({forest[i][j],i,j});
            }
        }

        sort(target_trees.begin(),target_trees.end());
        int no_of_trees = target_trees.size();
        int totalsteps=0;
        int startX=0;
        int startY=0;

        for(int t=0;t<no_of_trees;t++)
        {
     
            auto[height,targX,targY] = target_trees[t];
            int steps_to_reach_target_tree = stepsToCutNextTree(startX, startY, n, m, forest, targX, targY);
            if(steps_to_reach_target_tree == -1)
                return -1;
            
            totalsteps+=steps_to_reach_target_tree;
            startX=targX;
            startY=targY;
        }

        return totalsteps;
    }
};
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