Problem Description

You are given a forest represented as an m x n matrix where you need to cut down all trees for a golf event. Each cell in the matrix has the following meanings:

• 0: The cell is blocked and cannot be walked through
• 1: An empty cell that you can walk through
• A number greater than 1: A tree with that specific height that you can walk through

You start at position (0, 0) and can move one step at a time in four directions: north, east, south, or west. You can only move to cells that are walkable (not 0).

The key constraint is that you must cut the trees in order from shortest to tallest. Each tree has a unique height, and when you cut a tree, that cell becomes 1 (empty).

Your task is to find the minimum number of steps needed to walk from the starting position to cut all trees in the required order. If it's impossible to reach all trees, return -1.

For example, if there are trees with heights 2, 3, and 4, you must:

1. First walk from (0, 0) to the tree with height 2 and cut it
2. Then walk from that position to the tree with height 3 and cut it
3. Finally walk to the tree with height 4 and cut it

The total steps would be the sum of steps for each journey between consecutive trees.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The forest grid is essentially a graph where each cell is a node and adjacent walkable cells are connected by edges. We need to find paths between different positions in this grid.

Is it a tree?

• No: The grid structure is not a tree. It's a 2D grid graph where cycles can exist (you can walk in circles).

Is the problem related to directed acyclic graphs?

• No: The grid allows movement in all four directions, creating an undirected graph with potential cycles.

Is the problem related to shortest paths?

• Yes: We need to find the minimum number of steps (shortest path) from one position to each tree in sequence. The core challenge is finding the shortest path from our current position to the next tree we need to cut.

Is the graph Weighted?

• No: Each step from one cell to an adjacent cell has the same cost (1 step). All edges have uniform weight.

BFS

• Yes: Since we need shortest paths in an unweighted graph, BFS is the appropriate algorithm.

Conclusion: The flowchart correctly leads us to use BFS (Breadth-First Search) for this problem. We need to:

1. Sort all trees by height
2. Use BFS to find the shortest path from the starting position to the first tree
3. Use BFS again to find the shortest path from each tree to the next tree in the sorted order
4. Sum up all the shortest path distances

The solution implements BFS with an A* heuristic (using Manhattan distance) to optimize the search, but the core algorithm remains BFS for finding shortest paths in an unweighted grid graph.

Intuition

The key insight is to break down this problem into smaller, manageable pieces. We're not trying to find one optimal path that visits all trees - instead, we need to visit trees in a specific order (shortest to tallest).

Think of it like running errands in a city where you must visit stores in a specific order based on their opening times. You can't optimize the overall route freely; you must go to the earliest-opening store first, then the next, and so on.

This constraint naturally leads us to a greedy approach:

1. First, identify all trees and sort them by height
2. Visit them one by one in that sorted order
3. For each transition (current position → next tree), find the shortest path

Why BFS for each segment? Since we can move one step at a time in four directions (like moving on a chess board), and each move costs the same (1 step), we're dealing with an unweighted graph problem. BFS guarantees the shortest path in such scenarios because it explores all positions at distance d before exploring positions at distance d+1.

The solution cleverly optimizes the BFS with an A* heuristic using Manhattan distance abs(i - x) + abs(j - y). This helps prioritize exploring cells that are geometrically closer to the target, making the search more efficient. The Manhattan distance provides a lower bound on the actual distance since we can only move horizontally or vertically.

The overall strategy is:

• Decompose the problem: Instead of finding one complex path, find multiple simple shortest paths
• Order matters: We must respect the height ordering constraint
• Sum of local optimums: The minimum total steps equals the sum of minimum steps for each segment

If any tree is unreachable (BFS returns -1), the entire task becomes impossible, so we return -1.

Learn more about Breadth-First Search and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows a systematic approach to solve this problem:

1. Tree Collection and Sorting

trees = [(forest[i][j], i, j) for i in range(m) for j in range(n) if forest[i][j] > 1]
trees.sort()

We first scan the entire grid to find all trees (cells with value > 1), storing them as tuples (height, row, col). Sorting this list ensures we process trees in ascending order of height.

2. A Enhanced BFS Implementation* The bfs(i, j, x, y) function finds the shortest path from position (i, j) to target (x, y):

• Priority Queue with Heuristic: Uses a min-heap where entries are (f_score, row, col)

  • f_score = actual_distance + heuristic
  • The heuristic f(i, j, x, y) = abs(i - x) + abs(j - y) is the Manhattan distance

• Distance Tracking:

  dist = {i * n + j: 0}

  Maps each cell to its shortest known distance from the start. We use row * n + col as a unique identifier for each cell to avoid using tuples as dictionary keys (more efficient).

• Exploration Logic:

  for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
      c, d = i + a, j + b

  Explores all four directions from the current position.

• Path Validation: Only considers cells that are:

  • Within grid boundaries: 0 <= c < m and 0 <= d < n
  • Walkable: forest[c][d] > 0

• Distance Update: Updates the distance to a neighbor only if:

  • It hasn't been visited yet, OR
  • A shorter path is found: dist[c * n + d] > step + 1

3. Main Algorithm Flow

i = j = 0  # Start at (0, 0)
ans = 0    # Total steps counter

for _, x, y in trees:
    t = bfs(i, j, x, y)
    if t == -1:
        return -1  # Tree unreachable
    ans += t
    i, j = x, y  # Update current position

The algorithm:

1. Starts at (0, 0)
2. For each tree in height order:
   • Finds the shortest path from current position to the tree
   • If any tree is unreachable, returns -1
   • Adds the path length to the total
   • Updates the current position to the tree's location

Key Optimizations:

• A Heuristic*: The Manhattan distance heuristic guides the search toward the target, reducing unnecessary exploration
• Early Termination: Returns immediately when the target is reached
• Efficient Cell Indexing: Uses row * n + col instead of tuple keys for better performance

The time complexity is O(m²n²) in the worst case (for each tree, we might explore the entire grid), where m and n are the grid dimensions. The space complexity is O(mn) for the distance dictionary and priority queue.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Forest grid:
[[1, 2, 3],
 [0, 0, 4],
 [7, 6, 5]]

Starting position: (0, 0)

Step 1: Identify and sort trees

• Trees found: (2, 0, 1), (3, 0, 2), (4, 1, 2), (5, 2, 2), (6, 2, 1), (7, 2, 0)
• After sorting by height: [2, 3, 4, 5, 6, 7]

Step 2: Visit trees in order

Journey 1: (0,0) → Tree 2 at (0,1)

• Direct path: Right 1 step
• Steps: 1
• Current position: (0,1)

Journey 2: (0,1) → Tree 3 at (0,2)

• Direct path: Right 1 step
• Steps: 1
• Current position: (0,2)

Journey 3: (0,2) → Tree 4 at (1,2)

• Direct path blocked by cell (1,1) which is 0
• BFS finds: Can't go directly down
• Alternative path: Up is blocked (out of bounds), Left then around doesn't work
• Actually Tree 4 is reachable: Down 1 step (since (1,2) is walkable)
• Steps: 1
• Current position: (1,2)

Journey 4: (1,2) → Tree 5 at (2,2)

• Direct path: Down 1 step
• Steps: 1
• Current position: (2,2)

Journey 5: (2,2) → Tree 6 at (2,1)

• Direct path: Left 1 step
• Steps: 1
• Current position: (2,1)

Journey 6: (2,1) → Tree 7 at (2,0)

• Direct path: Left 1 step
• Steps: 1
• Current position: (2,0)

Total steps: 1 + 1 + 1 + 1 + 1 + 1 = 6

The A* optimization helps by prioritizing exploration toward the target. For instance, when finding a path from (0,1) to (0,2), the algorithm will immediately explore (0,2) since it has the lowest f_score:

• f_score = distance_traveled(1) + manhattan_distance(0) = 1

This prevents unnecessary exploration of other cells, making the search more efficient than standard BFS.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m²n² * (m + n))

The algorithm consists of finding paths between consecutive trees in sorted order:

• Sorting trees takes O(mnlog(mn)) where there are at most m*n trees
• For each tree (at most m*n trees), we perform an A* search using BFS
• Each A* search in worst case visits all m*n cells
• For each cell visited, we perform heap operations which take O(log(mn)) in worst case
• The heap can contain at most m*n elements
• Total for all BFS calls: O(mn * mn * log(mn)) = O(m²n²log(mn))

Since log(mn) < m + n for reasonable grid sizes, the overall complexity is O(m²n² * (m + n)).

Space Complexity: O(mn)

The space usage includes:

• trees list storing at most m*n tree positions: O(mn)
• In each BFS call:
  • Priority queue q can hold at most m*n elements: O(mn)
  • Dictionary dist can store at most m*n entries: O(mn)
• The space for BFS is reused for each search, not accumulated

Therefore, the total space complexity is O(mn).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Using Standard BFS Instead of A for Large Grids*

A common pitfall is implementing a standard BFS without the heuristic optimization. While BFS guarantees the shortest path, it explores cells uniformly in all directions, which can be inefficient for large grids.

Problem:

# Standard BFS without heuristic
queue = deque([(start_row, start_col, 0)])
visited = set()

Solution: Use A* with Manhattan distance heuristic to guide the search toward the target:

priority_queue = [(heuristic(start_row, start_col, target_row, target_col), 
                  start_row, start_col)]

2. Incorrect Distance Tracking Leading to Suboptimal Paths

A critical pitfall is not properly tracking or updating distances when a shorter path is found. This can happen when using a simple visited set instead of tracking actual distances.

Problem:

# Wrong: Only tracking visited cells
visited.add((row, col))
if (next_row, next_col) in visited:
    continue  # This prevents finding shorter paths

Solution: Track actual distances and update when finding shorter paths:

if (next_index not in distances or 
    distances[next_index] > next_distance):
    distances[next_index] = next_distance
    # Add to queue for further exploration

3. Forgetting to Handle the Starting Position (0, 0) Being Blocked

If the starting cell forest[0][0] = 0, the entire problem becomes unsolvable, but this edge case is often overlooked.

Problem:

# Not checking if start is valid
current_row = current_col = 0
# Directly starts processing without validation

Solution: Add validation at the beginning:

if forest[0][0] == 0:
    return -1  # Cannot start from a blocked cell

4. Using Tuples as Dictionary Keys (Performance Issue)

Using coordinate tuples as dictionary keys can be slower than using flattened indices, especially for large grids.

Problem:

# Slower approach with tuple keys
distances = {(start_row, start_col): 0}
if (next_row, next_col) not in distances:
    distances[(next_row, next_col)] = next_distance

Solution: Use flattened index for better performance:

# Faster with integer keys
distances = {start_row * cols + start_col: 0}
next_index = next_row * cols + next_col
if next_index not in distances:
    distances[next_index] = next_distance

5. Not Handling the Case Where Starting Position Has a Tree

If forest[0][0] > 1, it's a tree that needs to be cut, but since we start there, we need to ensure it's included in our tree list and handled correctly.

Problem:

# Might miss counting steps if (0,0) has a tree
for _, target_row, target_col in trees:
    # If first tree is at (0,0), find_shortest_path returns 0
    # but we still need to process it

Solution: The current implementation handles this correctly by:

• Including all trees in the sorted list regardless of position
• Starting path-finding from (0,0) to the first tree (even if it's at (0,0), returning 0 steps)
• Properly updating position after each tree

6. Integer Overflow in Cell Index Calculation

For very large grids, the flattened index row * cols + col might cause issues if not handled carefully.

Problem:

# For large m and n, this might overflow in some languages
next_index = next_row * cols + next_col

Solution: In Python, this isn't an issue due to arbitrary precision integers, but in other languages, consider:

• Using long/int64 types
• Or stick with tuple keys if grid size is extremely large
• Add bounds checking: assert rows * cols < sys.maxsize