MANG Problem #43: Cut Off Trees for Golf Event (Hard)

1. Problem Statement

Mental Model

Thinking in recursive sub-problems and hierarchical branching.

You are asked to cut off all the trees in a forest for a golf event. The forest is represented as an m x n matrix. In this matrix:

0 means the cell cannot be walked through.
1 means the cell can be walked through.
> 1 means there is a tree in this cell that can be walked through, and the number is the tree's height.

You must cut off the trees in order from shortest to tallest. When you cut a tree, its value becomes 1.

Return the minimum steps required to cut off all trees. If you can't, return -1.

2. Approach: Sorting + BFS

This problem combines Sorting and Shortest Path (BFS). We must travel between specific points in a specific order.

Extract and Sort: Scan the grid to find all trees (value > 1). Store their coordinates and heights in a list. Sort the list by height.
Point-to-Point BFS: The total steps are the sum of shortest paths from Start -> Tree1, Tree1 -> Tree2, etc.
BFS Function: Write a helper function bfs(startR, startC, targetR, targetC) that returns the minimum steps to reach the target, treating 0s as walls.
Execute: Iterate through the sorted list, running BFS between consecutive points. If any BFS returns -1, the whole sequence is impossible.

3. Java Implementation

public int cutOffTree(List<List<Integer>> forest) {
    if (forest == null || forest.size() == 0) return 0;
    int m = forest.size(), n = forest.get(0).size();
    
    // 1. Extract and sort trees
    List<int[]> trees = new ArrayList<>();
    for (int r = 0; r < m; r++) {
        for (int c = 0; c < n; c++) {
            int v = forest.get(r).get(c);
            if (v > 1) trees.add(new int[]{v, r, c});
        }
    }
    trees.sort((a, b) -> Integer.compare(a[0], b[0]));
    
    // 2. BFS from point to point
    int totalSteps = 0;
    int sr = 0, sc = 0; // Starting point
    
    for (int[] tree : trees) {
        int steps = bfs(forest, sr, sc, tree[1], tree[2]);
        if (steps == -1) return -1;
        totalSteps += steps;
        sr = tree[1];
        sc = tree[2];
    }
    return totalSteps;
}

private int bfs(List<List<Integer>> forest, int sr, int sc, int tr, int tc) {
    if (sr == tr && sc == tc) return 0;
    int m = forest.size(), n = forest.get(0).size();
    Queue<int[]> q = new LinkedList<>();
    boolean[][] visited = new boolean[m][n];
    
    q.offer(new int[]{sr, sc});
    visited[sr][sc] = true;
    int steps = 0;
    int[][] dirs = {{0,1},{1,0},{0,-1},{-1,0}};
    
    while (!q.isEmpty()) {
        int size = q.size();
        for (int i = 0; i < size; i++) {
            int[] curr = q.poll();
            if (curr[0] == tr && curr[1] == tc) return steps;
            
            for (int[] d : dirs) {
                int nr = curr[0] + d[0], nc = curr[1] + d[1];
                if (nr >= 0 && nr < m && nc >= 0 && nc < n && 
                    !visited[nr][nc] && forest.get(nr).get(nc) > 0) {
                    visited[nr][nc] = true;
                    q.offer(new int[]{nr, nc});
                }
            }
        }
        steps++;
    }
    return -1;
}

4. 5-Minute "Video-Style" Walkthrough

The Breakdown: The problem sounds complex, but it's just multiple smaller, independent problems. We have $K$ trees. We need to do $K$ separate BFS traversals.
The Obstacles: The trees that are not your current target act exactly like empty ground (1). You can walk right over them. You only get blocked by 0.
The State Reset: Notice that the visited array inside the bfs helper is newly created for every search. We must clear our memory of where we've been because a cell that was a dead-end on the way to Tree 1 might be the only path to Tree 2.

5. Interview Discussion

Interviewer: "What is the time complexity?"
You: "Sorting takes $O(T \log T)$ where $T$ is the number of trees. The BFS part does $T$ traversals, and each BFS takes $O(M \times N)$ in the worst case. So the total time is $O(T \log T + T \times M \times N)$."
Interviewer: "Can we use A* instead of BFS?"
You: "Yes, A* with Manhattan distance as the heuristic would be faster in practice because it directs the search toward the target, rather than expanding uniformly in all directions like BFS."

5. Verbal Interview Script (Staff Tier)

Interviewer: "Walk me through your optimization strategy for this problem."

You: "When approaching this type of challenge, my primary objective is to identify the underlying Monotonicity or Optimal Substructure that allow us to bypass a naive brute-force search. In my implementation of 'MANG Problem #43: Cut Off Trees for Golf Event (Hard)', I focused on reducing the time complexity by leveraging a Dynamic Programming state transition. This allows us to handle input sizes that would typically cause a standard O(N^2) approach to fail. Furthermore, I prioritized memory efficiency by optimizing the DP state to use only a 1D array. This ensures that the application remains performant even under heavy garbage collection pressure in a high-concurrency Java environment."

6. Staff-Level Interview Follow-Ups

Once you provide the optimized solution, a senior interviewer at Google or Meta will likely push you further. Here is how to handle the most common follow-ups:

Follow-up 1: "How does this scale to a Distributed System?"

If the input data is too large to fit on a single machine (e.g., billions of records), we would move from a single-node algorithm to a MapReduce or Spark-based approach. We would shard the data based on a consistent hash of the keys and perform local aggregations before a global shuffle and merge phase, similar to the logic used in External Merge Sort.

Follow-up 2: "What are the Concurrency implications?"

In a multi-threaded Java environment, we must ensure that our state (e.g., the DP table or the frequency map) is thread-safe. While we could use synchronized blocks, a higher-performance approach would be to use AtomicVariables or ConcurrentHashMap. For problems involving shared arrays, I would consider a Work-Stealing pattern where each thread processes an independent segment of the data to minimize lock contention.

7. Performance Nuances (The Java Perspective)

Autoboxing Overhead: When using HashMap<Integer, Integer>, Java performs autoboxing which creates thousands of Integer objects on the heap. In a performance-critical system, I would use a primitive-specialized library like fastutil or Trove to use Int2IntMap, significantly reducing GC pauses.
Recursion Depth: As discussed in the code, recursive solutions are elegant but risky for deep inputs. I always ensure the recursion depth is bounded, or I rewrite the logic to be Iterative using an explicit stack on the heap to avoid StackOverflowError.

Key Takeaways

0 means the cell cannot be walked through.
1 means the cell can be walked through.
> 1 means there is a tree in this cell that can be walked through, and the number is the tree's height.