675. Cut Off Trees for Golf Event


Problem Description

In this problem, we are given a matrix representing a forest where each cell can either be empty, blocked, or contain a tree with a certain height. The value '0' indicates a cell that cannot be walked through, the value '1' represents an empty cell that can be walked through, and any value above '1' is a tree with its height represented by that number. We can walk through the cells with trees, and upon doing so, we can choose to cut those trees down.

The primary objective is to cut all the trees in ascending order of their height and track the number of steps it would take to do so. We start from the top-left corner of the matrix (coordinates (0, 0)) and can move in the four cardinal directions: north, east, south, or west. Whenever a tree is cut down, its cell turns into an empty cell ('1'). If it is impossible to cut down all the trees, we have to return '-1'. Our goal is to determine the minimum number of steps required to achieve this task, given that we can only walk through non-zero cells and must cut down trees in order from the shortest to the tallest.

Intuition

To solve this problem, we need to find a way to navigate through the forest, cutting down trees in the specified order, and accumulating the number of steps taken to move from one tree to the next. The steps to approach the solution are as follows:

  1. Identify and Sort Trees by Height: We need to traverse all the cells in the matrix and identify the tree cells (values greater than '1'). We then sort these tree cells based on their height since we need to cut them in ascending order of their height.

  2. Calculate Minimum Distance Using BFS: We can use Breadth-First Search (BFS) to find the shortest path from one tree to the next. This algorithm allows us to explore the nearest neighbors of a cell first before moving on to cells that are further away, and thus we can find the shortest possible path step-by-step.

  3. Optimizing BFS with a Priority Queue (Heuristic): To optimize our BFS algorithm, we can use a priority queue (in Python, we use a min-heap). We add a heuristic function f(i, j, x, y) which is the Manhattan distance between the current cell (i, j) and the target cell (x, y). Along with the number of steps dist, this will decide the order of cells to be explored. The heap keeps the potential steps ordered by the estimated distance, so that the nearest cells are explored first.

  4. Perform BFS For Each Tree: Starting from the initial point (0, 0), we perform BFS for each tree in sorted order to get the minimum steps required to reach that tree. If at any point, a tree cannot be reached, we return '-1' as the answer.

  5. Summing Steps and Cutting Trees: As we move from one tree to the next, we keep adding the steps returned by BFS to our total step count. When a tree is cut, we must update the cell value to '1' to reflect that the cell is now empty and can be walked through for future paths.

  6. Return Total Steps: After all trees have been cut, we return the total number of steps taken as the solution.

By implementing the above approach, the cutOffTree function is designed to provide the minimum steps required to cut off all trees in the correct order, which is crucial for solving this LeetCode problem efficiently.

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

Solution Approach

The solution implements a combination of sorting, breadth-first search (BFS), and the use of a priority queue to find the minimum number of steps required to cut all trees in ascending order of their height.

Here's a breakdown of the implementation details for cutOffTree, which includes algorithms, data structures, or patterns:

  1. Initialize Variables: The forest dimensions m and n are determined by the length of the rows and columns of the input matrix. The starting point for the process is set to (0, 0).

  2. Tree Identification and Sorting: A list of tuples trees is created where each tuple contains the height of the tree and its coordinates (height, x, y). This list is populated by iterating over all cells in the forest matrix, and added to the list are cells with a value greater than 1 (tree cells). After collecting all tree cells, the list is sorted by the tree height to ensure that trees are cut in the correct order.

  3. Breadth-First Search (BFS):

    • The BFS function named bfs takes the current position (i, j) and the target position (x, y) as parameters.
    • A priority queue q is created, initially containing a tuple with the heuristic distance, the current position's coordinates, and the starting step count, 0.
    • A dictionary dist is used to track the shortest distance from the starting point to any cell (i, j).
    • While the queue is not empty, the cell with the smallest heuristic distance (estimated distance to the target) is popped, ensuring that cells closer to the target are visited first.
    • For every adjacent cell (a, b), we check if we can move into that cell (it's inside the grid and not a '0'). If it's a valid move and either not visited or offers a shorter path, update dist with the new distance and push the updated distance along with the heuristic into the queue.
  4. Iterating Over Trees:

    • Start from the initial position (i, j) = (0, 0) and iterate over all trees in the sorted trees list.
    • For each tree, call the bfs function to find the shortest path from the current position to the tree's position (x, y). The function returns the number of steps to reach the tree or -1 if the tree isn't reachable.
    • If a tree is unreachable (bfs returns -1), the function returns -1 immediately as it's impossible to cut all trees.
    • The steps returned from bfs for each successful tree cutting are accumulated in the variable ans. Upon each successful cut, the new current position is set to the tree's coordinates.
  5. Returning the Result: Once all trees are visited and cut, the total number of minimum steps accumulated in ans is returned.

By implementing these steps, we use BFS to calculate the minimum steps needed between each consecutive pair of trees and keep a cumulative count. The use of a priority queue with the Manhattan distance heuristic helps in traversing the forest efficiently. This gives us an optimized means to handle the problem's constraints and find the solution using Python's structures and algorithms.

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Example Walkthrough

Let's consider a small 3x3 forest matrix forest as our example:

1[
2    [1, 3, 0],
3    [0, 2, 1],
4    [1, 1, 4]
5]

Here, (1,1) is an empty cell, (0,0) is a blocked cell, 3, 2, and 4 are trees with heights mentioned.

Initialization

  • m = 3 (rows)
  • n = 3 (columns)
  • Start at (0, 0)

Tree Identification and Sorting

  • We identify the cells with trees:
    • 3 at (0, 1)
    • 2 at (1, 1)
    • 4 at (2, 2)
  • The tree list, unordered, is: [(3, 0, 1), (2, 1, 1), (4, 2, 2)]
  • After sorting, it becomes: [(2, 1, 1), (3, 0, 1), (4, 2, 2)]

Breadth-First Search (BFS)

  • Start BFS from (0, 0) to reach the first tree (2, 1, 1).
  • Since (0, 0) is blocked, move to an adjacent cell (0, 1) where tree 3 is located. However, we need to cut 2 first.
  • Because the direct path to (2, 1, 1) is blocked, bfs calculates a detour: (0, 1) -> (0, 2) -> (1, 2) -> (1, 1) with 3 steps.

Optimizing BFS and Moving to Next Trees

  • After cutting down tree 2, our current position is (1, 1).
  • Next, we calculate the path to tree 3 at (0, 1).
  • BFS reveals a single step is needed: (1, 1) -> (0, 1).
  • Cut tree 3, and the current position is (0, 1).
  • Finally, we find the shortest path to tree 4 at (2, 2), which takes 3 steps: (0, 1) -> (0, 2) -> (1, 2) -> (2, 2).

Summing Steps and Returning the Result

  • Total steps taken: 3 (to tree 2) + 1 (to tree 3) + 3 (to tree 4) = 7.
  • Since we were able to reach all trees, the result is 7.

By following these steps and utilizing the BFS with the priority queue and heuristic to find the shortest path through the forest, we can ensure that we cut all the trees in the required order using the minimum number of steps possible. If at any point a tree is unreachable, our implementation would return -1, representing an impossible task. In this case, all trees were reachable, and the minimum steps required to cut all trees were successfully calculated.

Solution Implementation

1from typing import List
2from heapq import heappop, heappush
3
4class Solution:
5    def cutOffTree(self, forest: List[List[int]]) -> int:
6        # Helper function to calculate Manhattan distance
7        def get_distance(i, j, x, y):
8            return abs(i - x) + abs(j - y)
9
10        # Perform a Breadth-First Search to find the shortest path
11        def bfs(start_i, start_j, dest_i, dest_j):
12            # Initialize the priority queue with the starting position
13            # and with initial distance as the heuristic
14            queue = [(get_distance(start_i, start_j, dest_i, dest_j), start_i, start_j)]
15            # Distance dictionary keeps track of steps from the start
16            distance = {start_i * column + start_j: 0}
17            while queue:
18                # Pop the position with lowest estimated cost
19                _, curr_i, curr_j = heappop(queue)
20                current_step = distance[curr_i * column + curr_j]
21                if (curr_i, curr_j) == (dest_i, dest_j):
22                    return current_step
23                # Check all four adjacent neighbors
24                for delta_i, delta_j in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
25                    next_i, next_j = curr_i + delta_i, curr_j + delta_j
26                    # Validate next position and check if it's a tree
27                    if 0 <= next_i < row and 0 <= next_j < column and forest[next_i][next_j] > 0:
28                        next_position_key = next_i * column + next_j
29                        if next_position_key not in distance or distance[next_position_key] > current_step + 1:
30                            distance[next_position_key] = current_step + 1
31                            heappush(queue, (distance[next_position_key] + get_distance(next_i, next_j, dest_i, dest_j), next_i, next_j))
32            return -1
33
34        row, column = len(forest), len(forest[0])
35        # Extract the trees and their positions and sort them based on height
36        trees = [(forest[i][j], i, j) for i in range(row) for j in range(column) if forest[i][j] > 1]
37        trees.sort()
38        current_i = current_j = total_steps = 0
39
40        # Iterate through all trees in sorted order
41        for _, target_i, target_j in trees:
42            # Get the distance to the next tree
43            steps_to_next_tree = bfs(current_i, current_j, target_i, target_j)
44            # If a tree can't be reached, return -1
45            if steps_to_next_tree == -1:
46                return -1
47            # Add the steps to the total steps and update current position
48            total_steps += steps_to_next_tree
49            current_i, current_j = target_i, target_j
50        return total_steps
51
1import java.util.*;
2
3class Solution {
4    private int[] distances = new int[3600]; // Pre-allocating space for distances.
5    private List<List<Integer>> forestGrid;  // The forest as a grid of trees.
6    private int rows;                       // Number of rows in the forest grid.
7    private int cols;                       // Number of columns in the forest grid.
8
9    public int cutOffTree(List<List<Integer>> forest) {
10        this.forestGrid = forest;
11        rows = forest.size();
12        cols = forest.get(0).size();
13        List<int[]> trees = new ArrayList<>();
14
15        // Collecting trees with height > 1 to cut them down in order of height.
16        for (int i = 0; i < rows; ++i) {
17            for (int j = 0; j < cols; ++j) {
18                int treeHeight = forest.get(i).get(j);
19                if (treeHeight > 1) {
20                    trees.add(new int[] {treeHeight, i * cols + j});
21                }
22            }
23        }
24
25        // Sorting trees by their heights because we need to cut trees in increasing order of height.
26        trees.sort(Comparator.comparingInt(a -> a[0]));
27
28        int steps = 0;  // The minimum number of steps to cut all trees.
29        int start = 0;  // Tracking the starting position of the worker (row * columns + column).
30
31        // Looping through the sorted trees to cut.
32        for (int[] tree : trees) {
33            int end = tree[1]; // Tree's position.
34            int stepToTree = bfs(start, end); // Using BFS to find the shortest path to the tree.
35            if (stepToTree == -1) {
36                return -1; // If tree is unreachable return -1.
37            }
38            steps += stepToTree; // Adding the steps to the final result.
39            start = end; // Update start position for the next tree.
40        }
41
42        return steps; // Returning the total number of steps needed to cut all trees.
43    }
44
45    private int bfs(int start, int end) {
46        PriorityQueue<int[]> queue = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])); // Priority Queue for selecting the node with minimum heuristic distance.
47        queue.offer(new int[] {calculateHeuristic(start, end), start});
48        Arrays.fill(distances, Integer.MAX_VALUE);
49        distances[start] = 0; // Starting point's distance is 0.
50
51        int[] directions = {-1, 0, 1, 0, -1}; // Directions: up, right, down, left.
52
53        // Process nodes in the queue.
54        while (!queue.isEmpty()) {
55            int[] node = queue.poll();
56            int currentState = node[1];
57            if (currentState == end) { // If end is reached.
58                return distances[currentState];
59            }
60
61            // Exploring adjacent squares in the forest.
62            for (int k = 0; k < 4; ++k) {
63                int newX = currentState / cols + directions[k];
64                int newY = currentState % cols + directions[k + 1];
65                if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && forestGrid.get(newX).get(newY) > 0) {
66                    if (distances[newX * cols + newY] > distances[currentState] + 1) {
67                        distances[newX * cols + newY] = distances[currentState] + 1;  // Update distance.
68                        queue.offer(new int[] {distances[newX * cols + newY] + calculateHeuristic(newX * cols + newY, end), newX * cols + newY});
69                    }
70                }
71            }
72        }
73
74        return -1; // If no path is found to the end, return -1;
75    }
76
77    // Heuristic function for A* search: computes the Manhattan distance between two points.
78    private int calculateHeuristic(int start, int end) {
79        int startX = start / cols;
80        int startY = start % cols;
81        int endX = end / cols;
82        int endY = end % cols;
83        return Math.abs(startX - endX) + Math.abs(startY - endY);
84    }
85}
86
1class Solution {
2public:
3    int rows; // "m" is renamed to "rows" to be more descriptive
4    int cols; // "n" is renamed to "cols" to be more descriptive
5    vector<int> distances; // "dist" is renamed to "distances"
6
7    int cutOffTree(vector<vector<int>>& forest) {
8        rows = forest.size();
9        cols = forest[0].size();
10        distances.resize(rows * cols); // Using rows * cols instead of a hardcoded value
11        vector<pair<int, int>> trees;
12
13        // Collect all the trees with a height greater than 1 and their locations
14        for (int i = 0; i < rows; ++i) {
15            for (int j = 0; j < cols; ++j) {
16                if (forest[i][j] > 1) {
17                    trees.emplace_back(forest[i][j], i * cols + j);
18                }
19            }
20        }
21
22        // Sort trees based on their height
23        sort(trees.begin(), trees.end());
24        int totalSteps = 0;
25        int currentPos = 0;
26
27        // Attempt to cut each tree in ascending order of height
28        for (auto& tree : trees) {
29            int targetPos = tree.second;
30            int stepsToTree = bfs(currentPos, targetPos, forest);
31            if (stepsToTree == -1) return -1; // If a tree is unreachable, return -1
32            totalSteps += stepsToTree;
33            currentPos = targetPos;
34        }
35
36        return totalSteps;
37    }
38
39    // Breadth-first search to find the shortest path to a tree
40    int bfs(int startPos, int endPos, vector<vector<int>>& forest) {
41        // PriorityQueue to get the node with the minimum distance from queue
42        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> minHeap;
43      
44        // Start from the initial position
45        minHeap.push({distanceHeuristic(startPos, endPos), startPos});
46      
47        // Initially, set all distances to infinity
48        fill(distances.begin(), distances.end(), INT_MAX);
49        distances[startPos] = 0;
50      
51        // Possible moves (up, right, down, left)
52        vector<int> directions = {-1, 0, 1, 0, -1};
53      
54        while (!minHeap.empty()) {
55            int currentPos = minHeap.top().second;
56            minHeap.pop();
57            if (currentPos == endPos) return distances[currentPos]; // Reached target
58          
59            // Explore all adjacent cells
60            for (int k = 0; k < 4; ++k) {
61                int newX = currentPos / cols + directions[k];
62                int newY = currentPos % cols + directions[k + 1];
63              
64                // If the cell is within bounds and not blocked
65                if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && forest[newX][newY] > 0) {
66                    int nextPos = newX * cols + newY;
67                    // If a shorter distance is found, update it
68                    if (distances[nextPos] > distances[currentPos] + 1) {
69                        distances[nextPos] = distances[currentPos] + 1;
70                        minHeap.push({distances[nextPos] + distanceHeuristic(nextPos, endPos), nextPos});
71                    }
72                }
73            }
74        }
75      
76        // If the path is not found, return -1
77        return -1;
78    }
79
80    // A heuristic function for estimating distance for priority queue (Manhattan distance)
81    int distanceHeuristic(int startPos, int endPos) {
82        int startX = startPos / cols;
83        int startY = startPos % cols;
84        int endX = endPos / cols;
85        int endY = endPos % cols;
86        return abs(startX - endX) + abs(startY - endY);
87    }
88};
89
1// Assuming that 'number' is a valid substitute for 'int' in the original C++ code
2// and we're using arrays and tuples to represent the vectors and pairs.
3
4// Define the number of rows and columns
5let rows: number;
6let cols: number;
7// Define the distances array
8let distances: number[];
9
10// Function to calculate the minimum steps to cut off all trees in a forest
11function cutOffTree(forest: number[][]): number {
12    rows = forest.length;
13    cols = forest[0].length;
14    // Initialize the distances with a size of rows * cols
15    distances = new Array(rows * cols);
16    // Collect trees with height greater than 1
17    const trees: [number, number][] = [];
18
19    // Collect all trees with a height greater than 1 and their locations
20    for (let i = 0; i < rows; ++i) {
21        for (let j = 0; j < cols; ++j) {
22            if (forest[i][j] > 1) {
23                trees.push([forest[i][j], i * cols + j]);
24            }
25        }
26    }
27
28    // Sort trees based on their height
29    trees.sort((a, b) => a[0] - b[0]);
30    let totalSteps = 0;
31    let currentPos = 0;
32
33    // Attempt to cut each tree in ascending order of height
34    for (const tree of trees) {
35        const targetPos = tree[1];
36        const stepsToTree = bfs(currentPos, targetPos, forest);
37        if (stepsToTree === -1) return -1; // If a tree is unreachable, return -1
38        totalSteps += stepsToTree;
39        currentPos = targetPos;
40    }
41
42    return totalSteps;
43}
44
45// Breadth-first search to find the shortest path to a tree
46function bfs(startPos: number, endPos: number, forest: number[][]): number {
47    // PriorityQueue is replaced with an array since JavaScript doesn't have a built-in PriorityQueue
48    type QueueElement = [number, number]; // Format: [heuristic+cost, position]
49    const minHeap: QueueElement[] = [];
50
51    // Start from the initial position
52    minHeap.push([distanceHeuristic(startPos, endPos), startPos]);
53
54    // Initially, set all distances to Infinity
55    distances.fill(Number.MAX_SAFE_INTEGER);
56    distances[startPos] = 0;
57
58    // Possible moves (up, right, down, left)
59    const directions = [-1, 0, 1, 0, -1];
60
61    while (minHeap.length > 0) {
62        // Since JavaScript lacks a built-in PriorityQueue, we manually sort and pop
63        minHeap.sort((a, b) => a[0] - b[0]);
64        const current = minHeap.shift();
65        if (!current) break;
66        const currentPos = current[1];
67      
68        if (currentPos === endPos) return distances[currentPos]; // Reached target
69
70        // Explore all adjacent cells
71        for (let k = 0; k < 4; ++k) {
72            const newX = Math.floor(currentPos / cols) + directions[k];
73            const newY = currentPos % cols + directions[k + 1];
74
75            // If the cell is within bounds and not blocked
76            if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && forest[newX][newY] > 0) {
77                const nextPos = newX * cols + newY;
78
79                // If a shorter distance is found, update it
80                if (distances[nextPos] > distances[currentPos] + 1) {
81                    distances[nextPos] = distances[currentPos] + 1;
82                    minHeap.push([distances[nextPos] + distanceHeuristic(nextPos, endPos), nextPos]);
83                }
84            }
85        }
86    }
87
88    // If the path is not found, return -1
89    return -1;
90}
91
92// A heuristic function for estimating distance for priority queue (Manhattan distance)
93function distanceHeuristic(startPos: number, endPos: number): number {
94    const startX = Math.floor(startPos / cols);
95    const startY = startPos % cols;
96    const endX = Math.floor(endPos / cols);
97    const endY = endPos % cols;
98    return Math.abs(startX - endX) + Math.abs(startY - endY);
99}
100

Time and Space Complexity

The given Python code performs a series of Breadth-First Searches (BFS) over a 2D grid to find the shortest path to cut trees in increasing order of their heights.

Time Complexity:

  • Sorting the trees: There are at most m*n trees, and sorting takes O(T log T) time where T is the number of trees. In the worst case, this is O(m*n log(m*n)).
  • BFS function (bfs): For each tree, we run a BFS. In the worst case, a BFS will visit every cell in the grid once. This is O(m*n) in the worst-case scenario for each tree.
  • Total BFS cost: Since BFS is executed for every tree, the worst-case time complexity for BFS across all trees is O(T * m * n) where T is the number of trees.

Overall, the total time complexity of the algorithm is O(m*n log(m*n) + T * m * n), which in practice is dominated by the BFS searches, so it can be approximated by O(T * m * n).

Space Complexity:

  • Queue/Heap in BFS: The heap can grow up to O(m*n) in space because, in the worst case, it might contain all the cells on the grid.
  • Visited dictionary (dist): The visited dictionary also stores information for each cell at most once, resulting in a space complexity of O(m*n).
  • Trees list: The list of trees sorted by height takes O(T) space.

Therefore, the total space complexity of the algorithm is the larger of O(m*n) (due to the heap and visited dictionary) and O(T) (due to the list of trees), which is O(m*n) since m*n is the upper bound on T.

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


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