<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>

// Build a Tree Cell Structure for simplicity sake //

struct Tree{
  
    public:
    
    int value;
    int x;
    int y;
    
};

// Helps to sort the trees on the basis of their values //

bool static comp(const Tree&x,const Tree&y){
    
    return x.value < y.value;
    
}

// 4-directional coordinates //

int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

int visited[52][52];

int bfs_traversal(int src1,int src2,int dest1,int dest2,vector<vector<int>>&forest){
    
    int m = forest.size();
    int n = forest[0].size();
    
    // Reset the visited array //
    
    for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
            visited[i][j] = 0;
        }
    }
    
    // Data structure widely used for BFS Traversal //
    
    queue<pair<int,int>>q;
    
    // Single Source BFS //
    
    q.push({src1,src2});
    
    int lvl = 0;
    
    while(!q.empty()){
        
        int sz = q.size();
        
        while(sz--){
            
            auto curr = q.front();
            q.pop();
            
            int x = curr.first;
            int y = curr.second;
            
            if(x==dest1 && y==dest2){
                return lvl;
            }
            
            if(visited[x][y]) continue;
            
            visited[x][y] = 1;
            
            for(int dir=0;dir<4;dir++){
                
                int nx = x + dx[dir];
                int ny = y + dy[dir];
                
                if(nx>=0 && nx<m && ny>=0 && ny<n && !visited[nx][ny] && forest[nx][ny]){
                    
                    q.push({nx,ny});
                    
                }
                
            }
            
        }
        
        lvl++;
        
    }
    
    return -1;
    
}

class Solution {
public:
    
    int cutOffTree(vector<vector<int>>& forest) {
        
        int m=forest.size();
        int n=forest[0].size();
        
        vector<Tree>vec;
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(forest[i][j]>1){
                    vec.push_back({forest[i][j],i,j});
                }
            }
        }
        
        // Starting Point - By default //
        
        vec.push_back({0,0,0});
        
        // Sort on the basis of the tree values //
        
        sort(vec.begin(),vec.end(),comp);
        
        int ans = 0;
        
        // BFS Traversal b/w adjacent trees //
        
        for(int i=0;i<vec.size()-1;i++){
            
            int curr = bfs_traversal(vec[i].x,vec[i].y,vec[i+1].x,vec[i+1].y,forest);
            
            // If there doesn't exist a single path b/w adjacent trees , then simple return -1 (invalid sequence) //
            
            if(curr<0){
                return -1;
            }
            
            ans += curr;
            
        }
        
        return ans;
        
    }
};

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