Leetcode #218: The Skyline Problem
In this guide, we solve Leetcode #218 The Skyline Problem in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return the skyline formed by these buildings collectively.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Binary Indexed Tree, Segment Tree, Array, Divide and Conquer, Ordered Set, Sorting, Line Sweep, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: buildings = [[2,9,10],[3,7,15],[5,12,12],[15,20,10],[19,24,8]]
Output: [[2,10],[3,15],[7,12],[12,0],[15,10],[20,8],[24,0]]
Explanation:
Figure A shows the buildings of the input.
Figure B shows the skyline formed by those buildings. The red points in figure B represent the key points in the output list.
Python Solution
from queue import PriorityQueue
class Solution:
def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
skys, lines, pq = [], [], PriorityQueue()
for build in buildings:
lines.extend([build[0], build[1]])
lines.sort()
city, n = 0, len(buildings)
for line in lines:
while city < n and buildings[city][0] <= line:
pq.put([-buildings[city][2], buildings[city][0], buildings[city][1]])
city += 1
while not pq.empty() and pq.queue[0][2] <= line:
pq.get()
high = 0
if not pq.empty():
high = -pq.queue[0][0]
if len(skys) > 0 and skys[-1][1] == high:
continue
skys.append([line, high])
return skys
Complexity
The time complexity is O(n log n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.