<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:
    // Directions for moving in the matrix: up, down, left, right
    vector<int> dr = {-1, 1, 0, 0};
    vector<int> dc = {0, 0, -1, 1};

    int cutOffTree(vector<vector<int>>& forest) {
        // List to store tree heights and their positions
        vector<vector<int>> trees;
        for (int r = 0; r < forest.size(); ++r) {
            for (int c = 0; c < forest[0].size(); ++c) {
                int v = forest[r][c];
                if (v > 1) {
                    trees.push_back({v, r, c}); // Add tree details to the list
                }
            }
        }

        // Sort the trees based on their heights
        sort(trees.begin(), trees.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] < b[0];
        });

        int ans = 0, sr = 0, sc = 0; // Start from the top-left corner
        for (const auto& tree : trees) {
            int d = dist(forest, sr, sc, tree[1], tree[2]);
            if (d < 0) {
                return -1; // If a tree is unreachable
            }
            ans += d; // Add the distance to the answer
            sr = tree[1]; // Update the current start row
            sc = tree[2]; // Update the current start column
        }
        return ans;
    }

    // Helper function to find the shortest path between two points
    int dist(const vector<vector<int>>& forest, int sr, int sc, int tr, int tc) {
        if (sr == tr && sc == tc) {
            return 0; // Already at the target
        }

        int R = forest.size(), C = forest[0].size();
        queue<pair<int, int>> q;
        q.push({sr, sc});
        vector<vector<bool>> visited(R, vector<bool>(C, false));
        visited[sr][sc] = true;
        int steps = 0;

        while (!q.empty()) {
            int size = q.size();
            ++steps;
            for (int i = 0; i < size; ++i) {
                auto [r, c] = q.front();
                q.pop();

                for (int j = 0; j < 4; ++j) {
                    int new_r = r + dr[j];
                    int new_c = c + dc[j];
                    if (new_r >= 0 && new_r < R && new_c >= 0 && new_c < C &&
                        !visited[new_r][new_c] && forest[new_r][new_c] != 0) {
                        if (new_r == tr && new_c == tc) {
                            return steps; // Found the shortest path
                        }
                        q.push({new_r, new_c});
                        visited[new_r][new_c] = true;
                    }
                }
            }
        }
        return -1; // Target is unreachable
    }
};

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