LeetCode #675 — HARD

Cut Off Trees for Golf Event

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 to cut it off.

You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes 1 (an empty cell).

Starting from the point (0, 0), return the minimum steps you need to walk to cut off all the trees. If you cannot cut off all the trees, return -1.

Note: 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.

Example 1:

Input: forest = [[1,2,3],[0,0,4],[7,6,5]]
Output: 6
Explanation: Following the path above allows you to cut off the trees from shortest to tallest in 6 steps.

Example 2:

Input: forest = [[1,2,3],[0,0,0],[7,6,5]]
Output: -1
Explanation: The trees in the bottom row cannot be accessed as the middle row is blocked.

Example 3:

Input: forest = [[2,3,4],[0,0,5],[8,7,6]]
Output: 6
Explanation: 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.

Constraints:

m == forest.length
n == forest[i].length
1 <= m, n <= 50
0 <= forest[i][j] <= 10⁹
Heights of all trees are distinct.

Step 01

Brute Force Baseline

Problem summary: 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 to cut it off. You must cut off the trees in order from shortest to tallest. When you cut off a tree, the value at its cell becomes 1 (an empty cell). Starting from the point (0, 0), return the minimum steps you need to walk to cut off all the trees. If you cannot cut off all the trees, return -1. Note: The input is generated such that no two trees have the same height, and there is at least one

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,2,3],[0,0,4],[7,6,5]]

Example 2

[[1,2,3],[0,0,0],[7,6,5]]

Example 3

[[2,3,4],[0,0,5],[8,7,6]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
class Solution {
    private int[] dist = new int[3600];
    private List<List<Integer>> forest;
    private int m;
    private int n;

    public int cutOffTree(List<List<Integer>> forest) {
        this.forest = forest;
        m = forest.size();
        n = forest.get(0).size();
        List<int[]> trees = new ArrayList<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (forest.get(i).get(j) > 1) {
                    trees.add(new int[] {forest.get(i).get(j), i * n + j});
                }
            }
        }
        trees.sort(Comparator.comparingInt(a -> a[0]));
        int ans = 0;
        int start = 0;
        for (int[] tree : trees) {
            int end = tree[1];
            int t = bfs(start, end);
            if (t == -1) {
                return -1;
            }
            ans += t;
            start = end;
        }
        return ans;
    }

    private int bfs(int start, int end) {
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        q.offer(new int[] {f(start, end), start});
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[start] = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            int state = q.poll()[1];
            if (state == end) {
                return dist[state];
            }
            for (int k = 0; k < 4; ++k) {
                int x = state / n + dirs[k];
                int y = state % n + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && forest.get(x).get(y) > 0) {
                    if (dist[x * n + y] > dist[state] + 1) {
                        dist[x * n + y] = dist[state] + 1;
                        q.offer(new int[] {dist[x * n + y] + f(x * n + y, end), x * n + y});
                    }
                }
            }
        }
        return -1;
    }

    private int f(int start, int end) {
        int a = start / n;
        int b = start % n;
        int c = end / n;
        int d = end % n;
        return Math.abs(a - c) + Math.abs(b - d);
    }
}

// Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
var dirs = [][]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}

type tree struct {
	height int
	pos    int
}

func cutOffTree(forest [][]int) int {
	row, col := len(forest), len(forest[0])

	bfs := func(start, end int) int {
		q := []int{start}
		vis := make(map[int]bool)
		vis[start] = true
		step := 0
		for n := len(q); n > 0; n = len(q) {
			for i := 0; i < n; i++ {
				state := q[0]
				q = q[1:]
				if state == end {
					return step
				}
				for k := 0; k < 4; k++ {
					x, y := state/col+dirs[k][0], state%col+dirs[k][1]
					nxt := x*col + y
					if x >= 0 && x < row && y >= 0 && y < col && forest[x][y] != 0 && !vis[nxt] {
						q = append(q, nxt)
						vis[nxt] = true
					}
				}
			}
			step++
		}
		return -1
	}

	var trees []tree
	for i := 0; i < row; i++ {
		for j := 0; j < col; j++ {
			if forest[i][j] > 1 {
				trees = append(trees, tree{forest[i][j], i*col + j})
			}
		}
	}
	sort.Slice(trees, func(i, j int) bool {
		return trees[i].height < trees[j].height
	})

	ans, start := 0, 0
	for _, t := range trees {
		end := t.pos
		step := bfs(start, end)
		if step == -1 {
			return -1
		}
		ans += step
		start = end
	}
	return ans
}

# Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
class Solution:
    def cutOffTree(self, forest: List[List[int]]) -> int:
        def f(i, j, x, y):
            return abs(i - x) + abs(j - y)

        def bfs(i, j, x, y):
            q = [(f(i, j, x, y), i, j)]
            dist = {i * n + j: 0}
            while q:
                _, i, j = heappop(q)
                step = dist[i * n + j]
                if (i, j) == (x, y):
                    return step
                for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
                    c, d = i + a, j + b
                    if 0 <= c < m and 0 <= d < n and forest[c][d] > 0:
                        if c * n + d not in dist or dist[c * n + d] > step + 1:
                            dist[c * n + d] = step + 1
                            heappush(q, (dist[c * n + d] + f(c, d, x, y), c, d))
            return -1

        m, n = len(forest), len(forest[0])
        trees = [
            (forest[i][j], i, j) for i in range(m) for j in range(n) if forest[i][j] > 1
        ]
        trees.sort()
        i = j = 0
        ans = 0
        for _, x, y in trees:
            t = bfs(i, j, x, y)
            if t == -1:
                return -1
            ans += t
            i, j = x, y
        return ans

// Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
use std::collections::HashSet;
use std::collections::VecDeque;

const DIRS: [[i32; 2]; 4] = [[-1, 0], [1, 0], [0, -1], [0, 1]];

impl Solution {
    pub fn cut_off_tree(forest: Vec<Vec<i32>>) -> i32 {
        let (row, col) = (forest.len() as i32, forest[0].len() as i32);

        let bfs = |start: i32, end: i32| -> i32 {
            let mut queue = VecDeque::new();
            let mut vis = HashSet::new();
            queue.push_back(start);
            vis.insert(start);
            let mut step = 0;
            while !queue.is_empty() {
                let n = queue.len();
                for _ in 0..n {
                    let state = queue.pop_front().unwrap();
                    if state == end {
                        return step;
                    }
                    for k in 0..4 {
                        let x = state / col + DIRS[k][0];
                        let y = (state % col) + DIRS[k][1];
                        let nxt = x * col + y;
                        if x >= 0
                            && x < row
                            && y >= 0
                            && y < col
                            && forest[x as usize][y as usize] != 0
                            && !vis.contains(&nxt)
                        {
                            queue.push_back(nxt);
                            vis.insert(nxt);
                        }
                    }
                }
                step += 1;
            }
            -1
        };

        let mut trees = Vec::new();
        for i in 0..row {
            for j in 0..col {
                let height = forest[i as usize][j as usize];
                if height > 1 {
                    trees.push((height, i * col + j));
                }
            }
        }
        trees.sort();

        let (mut ans, mut start) = (0, 0);
        for t in &trees {
            let end = t.1;
            let step = bfs(start, end);
            if step == -1 {
                return -1;
            }
            ans += step;
            start = end;
        }
        ans
    }
}

// Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #675: Cut Off Trees for Golf Event
// class Solution {
//     private int[] dist = new int[3600];
//     private List<List<Integer>> forest;
//     private int m;
//     private int n;
// 
//     public int cutOffTree(List<List<Integer>> forest) {
//         this.forest = forest;
//         m = forest.size();
//         n = forest.get(0).size();
//         List<int[]> trees = new ArrayList<>();
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (forest.get(i).get(j) > 1) {
//                     trees.add(new int[] {forest.get(i).get(j), i * n + j});
//                 }
//             }
//         }
//         trees.sort(Comparator.comparingInt(a -> a[0]));
//         int ans = 0;
//         int start = 0;
//         for (int[] tree : trees) {
//             int end = tree[1];
//             int t = bfs(start, end);
//             if (t == -1) {
//                 return -1;
//             }
//             ans += t;
//             start = end;
//         }
//         return ans;
//     }
// 
//     private int bfs(int start, int end) {
//         PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
//         q.offer(new int[] {f(start, end), start});
//         Arrays.fill(dist, Integer.MAX_VALUE);
//         dist[start] = 0;
//         int[] dirs = {-1, 0, 1, 0, -1};
//         while (!q.isEmpty()) {
//             int state = q.poll()[1];
//             if (state == end) {
//                 return dist[state];
//             }
//             for (int k = 0; k < 4; ++k) {
//                 int x = state / n + dirs[k];
//                 int y = state % n + dirs[k + 1];
//                 if (x >= 0 && x < m && y >= 0 && y < n && forest.get(x).get(y) > 0) {
//                     if (dist[x * n + y] > dist[state] + 1) {
//                         dist[x * n + y] = dist[state] + 1;
//                         q.offer(new int[] {dist[x * n + y] + f(x * n + y, end), x * n + y});
//                     }
//                 }
//             }
//         }
//         return -1;
//     }
// 
//     private int f(int start, int end) {
//         int a = start / n;
//         int b = start % n;
//         int c = end / n;
//         int d = end % n;
//         return Math.abs(a - c) + Math.abs(b - d);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.