Cut Off Trees for Golf Event
Time O(t * m * n) · Space O(m * n) · Official statement on LeetCode
Solutions
// Time: O(t * (logt + m * n)), t is the number of trees
// Space: O(t + m * n)
// Solution Reference:
// 1. https://discuss.leetcode.com/topic/103532/my-python-solution-inspired-by-a-algorithm/2
// 2. https://discuss.leetcode.com/topic/103562/python-solution-based-on-wufangjie-s-hadlock-s-algorithm
// 3. https://en.wikipedia.org/wiki/A*_search_algorithm
// 4. https://cg2010studio.files.wordpress.com/2011/12/dijkstra-vs-a-star.png
class Solution {
public:
int cutOffTree(vector<vector<int>>& forest) {
const auto m = forest.size(), n = forest[0].size();
priority_queue<pair<int, pair<int, int>>,
vector<pair<int, pair<int, int>>>,
greater<pair<int, pair<int, int>>> > min_heap;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (forest[i][j] > 1) {
min_heap.emplace(forest[i][j], make_pair(i, j));
}
}
}
pair<int, int> start;
int result = 0;
while (!min_heap.empty()) {
auto tree = min_heap.top(); min_heap.pop();
int step = minStep(forest, start, tree.second, m, n);
if (step < 0) {
return -1;
}
result += step;
start = tree.second;
}
return result;
}
private:
int minStep(const vector<vector<int>>& forest,
const pair<int, int>& start,
const pair<int, int>& end,
const int m, const int n) {
int min_steps = abs(start.first - end.first) + abs(start.second - end.second);
unordered_set<int> lookup;
vector<pair<int, int>> closer{start}, detour;
while (true) {
if (closer.empty()) { // cannot find a path in the closer expansions
if (detour.empty()) { // no other possible path
return -1;
}
// try other possible paths in detour expansions with extra 2-step cost
min_steps += 2;
swap(closer, detour);
}
int i, j;
tie(i, j) = closer.back(); closer.pop_back();
if (make_pair(i, j) == end) {
return min_steps;
}
if (!lookup.count(i * n + j)) {
lookup.emplace(i * n + j);
vector<pair<int, int>> expansions = {{i + 1, j}, {i - 1, j}, {i, j + 1}, {i, j - 1}};
for (const auto& expansion : expansions) {
int I, J;
tie(I, J) = expansion;
if (0 <= I && I < m && 0 <= J && J < n &&
forest[I][J] && !lookup.count(I * n + J)) {
bool is_closer = dot({I - i, J - j}, {end.first - i, end.second - j}) > 0;
is_closer ? closer.emplace_back(I, J) : detour.emplace_back(I, J);
}
}
}
}
return min_steps;
}
inline int dot(const pair<int, int>& a, const pair<int, int>& b) {
return a.first * b.first + a.second * b.second;
}
};
// Time: O(t * (logt + m * n)), t is the number of trees
// Space: O(t + m * n)
class Solution2 {
public:
int cutOffTree(vector<vector<int>>& forest) {
const auto m = forest.size(), n = forest[0].size();
priority_queue<pair<int, pair<int, int>>,
vector<pair<int, pair<int, int>>>,
greater<pair<int, pair<int, int>>> > min_heap;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (forest[i][j] > 1) {
min_heap.emplace(forest[i][j], make_pair(i, j));
}
}
}
pair<int, int> start;
int result = 0;
while (!min_heap.empty()) {
auto tree = min_heap.top(); min_heap.pop();
int step = minStep(forest, start, tree.second, m, n);
if (step < 0) {
return -1;
}
result += step;
start = tree.second;
}
return result;
}
private:
int minStep(const vector<vector<int>>& forest,
const pair<int, int>& start,
const pair<int, int>& end,
const int m, const int n) {
int min_steps = 0;
unordered_set<int> lookup;
queue<pair<int, int>> q;
q.emplace(start);
lookup.emplace(start.first * n + start.second);
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; ++i) {
auto curr = q.front(); q.pop();
if (curr == end) {
return min_steps;
}
static const vector<pair<int, int>> directions{{0, -1}, {0, 1},
{-1, 0}, {1, 0}};
for (const auto& direction : directions) {
int i = curr.first + direction.first;
int j = curr.second + direction.second;
if (i < 0 || i >= m || j < 0 || j >= n ||
!forest[i][j] || lookup.count(i * n + j)) {
continue;
}
q.emplace(i, j);
lookup.emplace(i * n + j);
}
}
++min_steps;
}
return -1;
}
};
Beginner Explanation
What is Cut Off Trees for Golf Event?
Cut Off Trees for Golf Event (LeetCode #675) is a Hard problem that primarily trains breadth first search.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with a search algorithm.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: A* Search Algorithm.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Cut Off Trees for Golf Event
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to a search algorithm.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(t * m * n)) and space (O(m * n)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(t * m * n) time and O(m * n) space.
Pattern focus: a search algorithm
Use the pattern as a checklist:
- a search algorithm — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(t * m * n) |
| Space | O(m * n) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Cut Off Trees for Golf Event
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for a search algorithm — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to a search algorithm:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: breadth first search.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Cut Off Trees for Golf Event in a second language (cpp, python).
- Drill 3–5 more problems tagged breadth first search.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the a search algorithm approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Cut Off Trees for Golf Event (#675) — Hard. Pattern: a search algorithm. Complexity: O(t * m * n) time / O(m * n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Cut Off Trees for Golf Event?+
The reference solutions aim for O(t * m * n) time and O(m * n) space. Always re-derive complexity from the code you write in the interview.
What pattern does Cut Off Trees for Golf Event use?+
It primarily maps to a search algorithm, within the broader topic of breadth first search.
Is Cut Off Trees for Golf Event good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/cut-off-trees-for-golf-event/