<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:
    const vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

// BFS to calculate shortest path from (startX, startY) to (targetX, targetY)
int bfs(vector<vector<int>>& forest, int startX, int startY, int targetX, int targetY) {
    int m = forest.size();
    int n = forest[0].size();
    
    // Edge case: if start == target, no need to move
    if (startX == targetX && startY == targetY) return 0;
    
    vector<vector<bool>> visited(m, vector<bool>(n, false));
    queue<pair<int, int>> q;
    q.push({startX, startY});
    visited[startX][startY] = true;
    int steps = 0;
    
    while (!q.empty()) {
        int size = q.size();
        steps++;
        
        for (int i = 0; i < size; i++) {
            auto [x, y] = q.front(); q.pop();
            
            // Explore all four possible directions
            for (auto& dir : directions) {
                int newX = x + dir.first;
                int newY = y + dir.second;
                
                if (newX >= 0 && newX < m && newY >= 0 && newY < n &&
                    !visited[newX][newY] && forest[newX][newY] != 0) {
                    if (newX == targetX && newY == targetY) {
                        return steps; // Reached the target tree
                    }
                    q.push({newX, newY});
                    visited[newX][newY] = true;
                }
            }
        }
    }
    
    // If we exit the while loop, the target is unreachable
    return -1;
}

int cutOffTree(vector<vector<int>>& forest) {
    int m = forest.size();
    int n = forest[0].size();
    
    // Step 1: Gather all trees and sort them by height
    vector<pair<int, pair<int, int>>> trees; // {height, {x, y}}
    
    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 by height
    sort(trees.begin(), trees.end());
    
    // Step 2: Traverse through the sorted trees and calculate total steps
    int totalSteps = 0;
    int startX = 0, startY = 0;
    
    for (const auto& tree : trees) {
        int targetX = tree.second.first;
        int targetY = tree.second.second;
        
        int steps = bfs(forest, startX, startY, targetX, targetY);
        if (steps == -1) {
            return -1; // If any tree is unreachable, return -1
        }
        
        totalSteps += steps;
        // Update the current position to the tree's position
        startX = targetX;
        startY = targetY;
        
        // Mark the tree as cut by setting its value to 1
        forest[targetX][targetY] = 1;
    }
    
    return totalSteps;
}

};
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