What is the time complexity of Cut Off Trees for Golf Event?

The optimal solution has a time complexity of O(m*n*log(m*n) + t*m*n).

What is the space complexity of Cut Off Trees for Golf Event?

The space complexity is O(m*n).

What data structures are used to solve Cut Off Trees for Golf Event?

This problem uses Array, Breadth-First Search, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Cut Off Trees for Golf Event?

Insight 1: The order of tree cutting is determined by their heights, requiring sorting to process them correctly. Insight 2: Breadth-First Search (BFS) is the optimal algorithm to find the shortest path between two points in a grid.

Which companies ask LeetCode 675. Cut Off Trees for Golf Event in interviews?

This problem is frequently asked at Flipkart.

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Cut Off Trees for Golf Event - LeetCode 675 Solution

Cut Off Trees for Golf Event - Complete Solution Guide

Cut Off Trees for Golf Event is LeetCode problem 675, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 represents an empty cell that can be walked through. A number greater than 1 represents a tree in a cell that can be walked through, and this number is the tree's height. 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

Detailed Explanation

The problem requires you to find the minimum number of steps to cut down trees in a forest represented by a 2D grid, starting from (0, 0). Trees are represented by values greater than 1, and you must cut them down in order of their height (shortest to tallest). 0 represents an obstacle, and 1 represents an empty cell. You can move in four directions (north, east, south, west). After cutting a tree, the cell's value becomes 1. If it's impossible to cut down all trees, return -1. The key constraint is that all tree heights are unique.

Solution Approach

The solution first identifies all tree locations and their heights. It then sorts the trees by height in ascending order. Starting from (0, 0), it iterates through the sorted list of trees. For each tree, it uses BFS to find the shortest path from the current location to the tree's location. If a path is found, the steps are added to the total steps, and the current location is updated to the tree's location. If no path is found to any tree, the function returns -1. The algorithm effectively simulates cutting down the trees in the required order and accumulating the steps needed to reach each tree.

Step-by-Step Algorithm

Step 1: Find all trees (cells with values > 1) in the forest and store their heights and coordinates in a list.
Step 2: Sort the list of trees by height in ascending order.
Step 3: Initialize the starting position to (0, 0) and the total steps to 0.
Step 4: Iterate through the sorted list of trees.
Step 5: For each tree, use BFS to find the shortest path from the current position to the tree's location.
Step 6: If BFS returns -1 (no path found), return -1 (impossible to cut all trees).
Step 7: If BFS returns a path length, add it to the total steps.
Step 8: Update the current position to the tree's location after 'cutting' the tree (implicitly; the forest itself is not modified in place).
Step 9: After processing all trees, return the total steps.

Key Insights

Insight 1: The order of tree cutting is determined by their heights, requiring sorting to process them correctly.
Insight 2: Breadth-First Search (BFS) is the optimal algorithm to find the shortest path between two points in a grid.
Insight 3: It's essential to handle cases where a path between two trees doesn't exist (return -1).

Complexity Analysis

Time Complexity: O(m*n*log(m*n) + t*m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Breadth-First Search, Heap (Priority Queue), Matrix.

Companies

Asked at: Flipkart.

Cut Off Trees for Golf Event | LeetCode 675

Hard

Array Breadth-First Search Heap (Priority Queue)Matrix

Time: O(m*n*log(m*n) + t*m*n)Space: O(m*n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int cutOffTree(List<List<Integer>> forest) {
        int m = forest.size();
        int n = forest.get(0).size();

        List<int[]> trees = new ArrayList<>();
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (forest.get(r).get(c) > 1) {
                    trees.add(new int[]{forest.get(r).get(c), r, c});
                }
            }
        }

        trees.sort(Comparator.comparingInt(a -> a[0]));

        class Pair {
            int row;
            int col;
            int steps;

            public Pair(int row, int col, int steps) {
                this.row = row;
                this.col = col;
                this.steps = steps;
            }
        }

        class Tuple {
            int r;
            int c;
            public Tuple(int r, int c) {
                this.r = r;
                this.c = c;
            }

            @Override
            public boolean equals(Object o) {
                if (this == o) return true;
                if (o == null || getClass() != o.getClass()) return false;
                Tuple tuple = (Tuple) o;
                return r == tuple.r && c == tuple.c;
            }

            @Override
            public int hashCode() {
                return Objects.hash(r, c);
            }
        }

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Problem Info

Description

Problem 675: Cut Off Trees for Golf Event - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Heap (Priority Queue)Matrix

Companies

Flipkart

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:675

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Find all trees (cells with values > 1) in the forest and store their heights and coordinates in a list.
2
Step 2: Sort the list of trees by height in ascending order.
3
Step 3: Initialize the starting position to (0, 0) and the total steps to 0.
4
Step 4: Iterate through the sorted list of trees.
5
Step 5: For each tree, use BFS to find the shortest path from the current position to the tree's location.
6
Step 6: If BFS returns -1 (no path found), return -1 (impossible to cut all trees).
7
Step 7: If BFS returns a path length, add it to the total steps.
8
Step 8: Update the current position to the tree's location after 'cutting' the tree (implicitly; the forest itself is not modified in place).
9
Step 9: After processing all trees, return the total steps.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*log(m*n) + t*m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is derived from the sorting and BFS operations. Sorting dominates when the number of trees is high. The space complexity is determined by the maximum size of data structures used during BFS and for storing tree locations.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A*

Instead of BFS, A* search can be used to find the shortest path. A* uses a heuristic function to guide the search, potentially improving performance in some cases where BFS explores many irrelevant nodes.

Time: O(t*m*n) in the worst case, but can be faster with a good heuristic.Space: O(m*n)

Trade-offs: A* can be faster than BFS with a good heuristic, especially in larger forests with fewer obstacles. However, a poorly chosen heuristic can make A* slower than BFS. BFS is generally simpler to implement.

✨

Key Insights

1Insight 1: The order of tree cutting is determined by their heights, requiring sorting to process them correctly.
2Insight 2: Breadth-First Search (BFS) is the optimal algorithm to find the shortest path between two points in a grid.
3Insight 3: It's essential to handle cases where a path between two trees doesn't exist (return -1).

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: The forest is empty or contains no trees with height > 1. The code correctly handles this by not adding any trees to the 'trees' list, resulting in an empty loop, and thus returning 0.
Edge case 2: There is no path between two trees. The BFS function will return -1, and the `cutOffTree` function will also return -1, indicating it is impossible to cut down all trees.
Edge case 3: Starting point is blocked (forest[0][0] == 0). The problem statement implies that forest[0][0] will always be walkable (value 1 or greater), thus it is implicitly handled without special code.
Edge case 4: The target tree is unreachable. BFS will explore all reachable cells and if the target is not reached, returns -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the trees by height before cutting them down. This will lead to incorrect results.

✗Common mistake 2: Using Depth-First Search (DFS) instead of BFS. DFS does not guarantee finding the shortest path.

✗Common mistake 3: Not handling the case where a tree is unreachable, causing an infinite loop or incorrect results.

✗Common mistake 4: Modifying the input 'forest' array in-place. While the problem says cells are updated to 1 after being cut, the given solutions do not modify the array because the shortest path algorithm only requires the tree locations, not the actual modification of the forest.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Robot path planning in a warehouse or factory. The algorithm can be used to find the shortest path for a robot to visit a set of locations in a specific order, avoiding obstacles.

🏦Real-world scenario 2: Delivery route optimization. The algorithm can be adapted to find the most efficient route for a delivery driver to visit a set of customers, minimizing travel time and distance.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Cut Off Trees for Golf Event"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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