<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 {
   private:
    vector<vector<int>> moves = {{0, 1},
                                 {0, -1},
                                 {1, 0},
                                 {-1, 0}};

    bool isValid(int x, int y, int m, int n) {
        return !(x < 0 || x >= m || y < 0 || y >= n);
    }

    int steps(int currX,
              int currY,
              int finalX,
              int finalY,
              vector<vector<int>>& forest,
              vector<vector<int>>& visited) {
        
        if (currX == finalX && currY == finalY) {
                            return 0;
                        }

        int m = forest.size(), n = forest[0].size();
        queue<pair<int, int>> bfsQ;

        bfsQ.push({currX, currY});
        visited[currX][currY] = 1;
        int step = 0;

        while (bfsQ.size()) {
            step++;
            int sz = bfsQ.size();

            for (int i = 0; i < sz; i++) {
                pair<int, int> top = bfsQ.front();
                bfsQ.pop();
                for (vector<int> &move : moves) {
                    int newX = top.first + move[0], newY = top.second + move[1];
                    if (isValid(newX, newY, m, n) && !visited[newX][newY] &&
                        forest[newX][newY] >= 1) {
                        visited[newX][newY] = 1;
                        if (newX == finalX && newY == finalY) {
                            return step;
                        }
                        bfsQ.push({newX, newY});
                    }
                }
            }
        }

        return -1;
    }

   public:
    int cutOffTree(vector<vector<int>>& forest) {
        int m = forest.size(), n = forest[0].size(), netSteps = 0;
        vector<vector<int>> trees, visited(m, vector<int>(n, 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], i, j});
            }
        }

        sort(trees.begin(), trees.end());

        int currX = 0, currY = 0;
        for (int i = 0; i < (int)trees.size(); i++) {
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++)
                    visited[i][j] = 0;
            }
            int req = steps(currX, currY, trees[i][1], trees[i][2], forest, visited);

            if (req == -1)
                return -1;

            netSteps += req;
            currX = trees[i][1], currY = trees[i][2];
            forest[currX][currY] = 1;
        }

        return netSteps;
    }
};
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