<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:
    int m, n;

    int heuristic(int x1, int y1, int x2, int y2) {
        return abs(x1 - x2) + abs(y1 - y2); //Manhattan distance
    }

    int nextStep(std::vector<std::vector<int>> &forest, int startRow, int startCol, int endRow, int endCol) {
        //A* Search, T-O(m*n), S-O(m*n)
        if (startRow == endRow && startCol == endCol) return 0;

        //Min-heap priority queue based on the estimated total cost (g + h)
        std::priority_queue<std::tuple<int, int, int>, std::vector<std::tuple<int, int, int>>, std::greater<>> pq;
        pq.push({0, startRow, startCol});

        std::vector<std::vector<bool>> visited(m, std::vector<bool>(n, false));
        visited[startRow][startCol] = true;

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

        while (!pq.empty()) {
            std::tuple<int, int, int> p = pq.top();//extract current g (steps) and the current position
            int steps = std::get<0>(p), row = std::get<1>(p),col = std::get<2>(p);
            pq.pop();

            for (auto &dir : directions) {
                int r = row + dir[0];
                int c = col + dir[1];
                if (r >= 0 && r < m && c >= 0 && c < n && !visited[r][c] && forest[r][c] > 0) {
                    if (r == endRow && c == endCol) return steps + 1;
                    visited[r][c] = true;
                    int h = heuristic(r, c, endRow, endCol);
                    pq.push({steps + 1, r, c}); // only push g (steps) onto the queue, not g+h
                }
            }
        }
        return -1;
    }

public:
    int cutOffTree(vector<vector<int>>& forest) {
        //Sorting then traversal of all trees then A* Search, T-O(k*m*n), S-O(m*n), k is the number of trees to be cut in the forest
        m = forest.size(), n = forest[0].size();
        std::vector<std::vector<int>> trees;
        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});//value, row, col


        std::sort(trees.begin(), trees.end());//by default sorts by value i.e. forest[i][j] here

        int stepsTaken = 0;
        for (int i = 0, currRow = 0, currCol = 0; i < trees.size(); ++i) {
            int steps = nextStep(forest, currRow, currCol, trees[i][1], trees[i][2]);
            if (steps == -1) return -1;
            stepsTaken += steps;
            currRow = trees[i][1];
            currCol = trees[i][2];
        }

        return stepsTaken;
    }
};
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