<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>

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
//memory can be enchanced if these are passed by refrence in functions
int n, m, x, y, dist[50 + 5][50 + 5], ret = 0;
set<int> st;
vector<vector<int>> v;

class Solution {
public:
    // if out of bounds from the array
    bool valid(int a, int b) {
        return (a >= 0 && a < n && b >= 0 && b < m && v[a][b]);
    }

    // Bfs shortest path algorithm
    // getting the shortest path from the current (i, j) to all valid nodes
    // time can be enhaced if we return after we hit the node we are looking for
    void bfs(int i, int j) {
        queue<pair<int, int>> q;
        q.emplace(i, j), dist[i][j] = 0;
        while (!q.empty()) {
            int ff = q.front().first, xx = q.front().second;
            q.pop();
            
            // looping through the direction array
            for (int k = 0; k < 4; ++k) {
                int ni = ff + dx[k], nj = xx + dy[k];

                if (valid(ni, nj) && dist[ff][xx] + 1 < dist[ni][nj]) {
                    q.emplace(ni, nj);
                    dist[ni][nj] = dist[ff][xx] + 1;
                }
            }
        }
    }

    int cutOffTree(vector<vector<int>> forest) {
        v = forest;
        if (!v[0][0]) return -1;
        ret = 0, x = 0, y = 0;
        st.clear();

        // inserting nodes in order through a set
        n = v.size(), m = v[0].size();
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                if (v[i][j]) st.insert(v[i][j]);

        for (auto s : st) {
            if (s == 1) continue;

            // reseting dist for shortest path algorithm
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < m; ++j) dist[i][j] = INT_MAX;

            bfs(x, y);
        
            int ni = x, nj = y;
            for (int k = 0; k < n; ++k)
                for (int t = 0; t < m; ++t)
                    if (v[k][t] == s && dist[k][t] != INT_MAX)
                        ret += dist[k][t], ni = k, nj = t;

            // having ni and nj, if they are the same after we loop
            // this means we could not reach the desired node
            if (ni == x && nj == y && s != *st.begin()) return -1;

            // starting the next bfs from here
            x = ni, y = nj;
        }
        return ret;
    }   
};
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