The Skyline Problem
Time O(nlogn) · Space O(n) · Official statement on LeetCode
Solutions
// Time: O(nlogn)
// Space: O(n)
// BST solution.
class Solution {
public:
enum {start, end, height};
struct Endpoint {
int height;
bool isStart;
};
vector<pair<int, int> > getSkyline(vector<vector<int> >& buildings) {
map<int, vector<Endpoint>> point_to_height; // Ordered, no duplicates.
for (const auto& building : buildings) {
point_to_height[building[start]].emplace_back(Endpoint{building[height], true});
point_to_height[building[end]].emplace_back(Endpoint{building[height], false});
}
vector<pair<int, int>> res;
map<int, int> height_to_count; // BST.
int curr_max = 0;
// Enumerate each point in increasing order.
for (const auto& kvp : point_to_height) {
const auto& point = kvp.first;
const auto& heights = kvp.second;
for (const auto& h : heights) {
if (h.isStart) {
++height_to_count[h.height];
} else {
--height_to_count[h.height];
if (height_to_count[h.height] == 0) {
height_to_count.erase(h.height);
}
}
}
if (height_to_count.empty() ||
curr_max != height_to_count.crbegin()->first) {
curr_max = height_to_count.empty() ?
0 : height_to_count.crbegin()->first;
res.emplace_back(point, curr_max);
}
}
return res;
}
};
// Time: O(nlogn)
// Space: O(n)
// Divide and conquer solution.
class Solution2 {
public:
enum {start, end, height};
vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) {
const auto intervals = ComputeSkylineInInterval(buildings, 0, buildings.size());
vector<pair<int, int>> res;
int last_end = -1;
for (const auto& interval : intervals) {
if (last_end != -1 && last_end < interval[start]) {
res.emplace_back(last_end, 0);
}
res.emplace_back(interval[start], interval[height]);
last_end = interval[end];
}
if (last_end != -1) {
res.emplace_back(last_end, 0);
}
return res;
}
// Divide and Conquer.
vector<vector<int>> ComputeSkylineInInterval(const vector<vector<int>>& buildings,
int left_endpoint, int right_endpoint) {
if (right_endpoint - left_endpoint <= 1) { // 0 or 1 skyline, just copy it.
return {buildings.cbegin() + left_endpoint,
buildings.cbegin() + right_endpoint};
}
int mid = left_endpoint + ((right_endpoint - left_endpoint) / 2);
auto left_skyline = ComputeSkylineInInterval(buildings, left_endpoint, mid);
auto right_skyline = ComputeSkylineInInterval(buildings, mid, right_endpoint);
return MergeSkylines(left_skyline, right_skyline);
}
// Merge Sort
vector<vector<int>> MergeSkylines(vector<vector<int>>& left_skyline, vector<vector<int>>& right_skyline) {
int i = 0, j = 0;
vector<vector<int>> merged;
while (i < left_skyline.size() && j < right_skyline.size()) {
if (left_skyline[i][end] < right_skyline[j][start]) {
merged.emplace_back(move(left_skyline[i++]));
} else if (right_skyline[j][end] < left_skyline[i][start]) {
merged.emplace_back(move(right_skyline[j++]));
} else if (left_skyline[i][start] <= right_skyline[j][start]) {
MergeIntersectSkylines(merged, left_skyline[i], i,
right_skyline[j], j);
} else { // left_skyline[i][start] > right_skyline[j][start].
MergeIntersectSkylines(merged, right_skyline[j], j,
left_skyline[i], i);
}
}
// Insert the remaining skylines.
merged.insert(merged.end(), left_skyline.begin() + i, left_skyline.end());
merged.insert(merged.end(), right_skyline.begin() + j, right_skyline.end());
return merged;
}
// a[start] <= b[start]
void MergeIntersectSkylines(vector<vector<int>>& merged, vector<int>& a, int& a_idx,
vector<int>& b, int& b_idx) {
if (a[end] <= b[end]) {
if (a[height] > b[height]) { // |aaa|
if (b[end] != a[end]) { // |abb|b
b[start] = a[end];
merged.emplace_back(move(a)), ++a_idx;
} else { // aaa
++b_idx; // abb
}
} else if (a[height] == b[height]) { // abb
b[start] = a[start], ++a_idx; // abb
} else { // a[height] < b[height].
if (a[start] != b[start]) { // bb
merged.emplace_back(move(vector<int>{a[start], b[start], a[height]})); // |a|bb
}
++a_idx;
}
} else { // a[end] > b[end].
if (a[height] >= b[height]) { // aaaa
++b_idx; // abba
} else {
// |bb|
// |a||bb|a
if (a[start] != b[start]) {
merged.emplace_back(move(vector<int>{a[start], b[start], a[height]}));
}
a[start] = b[end];
merged.emplace_back(move(b)), ++b_idx;
}
}
}
};
Beginner Explanation
What is The Skyline Problem?
The Skyline Problem (LeetCode #218) is a Hard problem that primarily trains sort.
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 sort.
- 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: Sort, BST.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for The Skyline Problem
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 sort.
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(nlogn)) and space (O(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(nlogn) time and O(n) space.
Pattern focus: sort
Use the pattern as a checklist:
- sort — 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(nlogn) |
| Space | O(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 The Skyline Problem
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for sort — 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 sort:
- 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: sort.
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 The Skyline Problem in a second language (cpp, python).
- Drill 3–5 more problems tagged sort.
- 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 sort 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
The Skyline Problem (#218) — Hard. Pattern: sort. Complexity: O(nlogn) time / O(n) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of The Skyline Problem?+
The reference solutions aim for O(nlogn) time and O(n) space. Always re-derive complexity from the code you write in the interview.
What pattern does The Skyline Problem use?+
It primarily maps to sort, within the broader topic of sort.
Is The Skyline Problem 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/the-skyline-problem/