<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) {

        // Given : Clear the trees in increasing order of height
        // constraints : n , m <= 50 = Larger Complexity Acceptable

        // Solution : To handle the contriant we are given that we have to cut them
        // in increasing order which provides us with solution itself

        // If gives tress are [ 10 , 12 , 11 ]
        // if i cut 10 next i have to cut will be 11 and i need to find a way to reach 11 
        // if possible find steps to it if not return -1

        vector<vector<int>> trees;
        for(int i = 0; i < forest.size(); i++){

            for(int j = 0; j < forest[0].size(); j++){

                if(forest[i][j] > 1){

                    trees.push_back({forest[i][j] , i , j});
                }
            }
            cout<<endl;
        }

        // sorting the trees according to their height
        sort(trees.begin() , trees.end());

        // how it gurantees minimum ? As bfs does a level order search so we will
        // surely get the closest step automatically

        // defining all valid moves in advance
        vector<pair<int,int>> moves = {{0 , 1} , {1 , 0} , { 0 , -1} , {-1 , 0}};

        // defining start points 
        int sx = 0 , sy = 0;
        int ans = 0;
        // now we will cut trees in order
        for(int i = 0; i < trees.size(); i++){
            
            // using breadth first search to find if next tree is reachable or not
            int cost = bfs(forest , sx , sy , trees[i][1] , trees[i][2] , moves);
            // if not reachable return -1
            if(cost == -1){
                
                return -1;
            }
            // if reachable add the 1 to cost
            ans += cost;
            // update the start points to new tree
            sx = trees[i][1];
            sy = trees[i][2];

        }

        return ans;
        
    }

    int bfs(vector<vector<int>> forest , int sx , int sy , int fx , int fy , vector<pair<int,int>> &moves){

        // if we are search for start point
        if(fx == sx && fy == sy) return 0;

        // now let's do a bfs 

        queue<pair<int,int>> q;
        q.push({sx , sy});
        int n = forest.size() , m = forest[0].size();
        vector<vector<int>> seen(n , vector<int>(m , 0));
        seen[sx][sy] = 1;
        int cost = 0;
        while(!q.empty()){

            int qsize = q.size();
            // iterating the current level
            for(int i = 0; i < qsize; i++){

                pair<int,int> point = q.front();
                q.pop();

                for(int j = 0 ; j < moves.size(); j++){

                    long int newx = point.first + moves[j].first;
                    long int newy = point.second + moves[j].second;

                    // skip if new point is not valid point
                    if(newx >= n || newy >= m || newx < 0 || newy < 0 ) continue;

                    // if new point is not obstacle and also not seen before the add it
                    if(forest[newx][newy] != 0 && seen[newx][newy] == 0){

                        q.push({newx , newy});
                        seen[newx][newy] = 1;

                        if(newx == fx && newy == fy){

                            return cost + 1;
                        }

                    }
    
                }
                
            }
            
            cost++;


        }

        // if not found return -1
        return -1;


    }
};
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