The Skyline Problem - Practice Coding Problems

The Skyline Problem - Problem

Array Divide and Conquer Binary Indexed Tree Segment Tree Line Sweep Sorting Heap (Priority Queue) Ordered Set Hard

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.

The geometric information of each building is given in the array buildings where buildings[i] = [left_i, right_i, height_i]:

left_i is the x coordinate of the left edge of the ith building.
right_i is the x coordinate of the right edge of the ith building.
height_i is the height of the ith building.

You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.

The skyline should be represented as a list of "key points" sorted by their x-coordinate in the form [[x1,y1],[x2,y2],...]. Each key point is the left endpoint of some horizontal segment in the skyline except the last point in the list, which always has a y-coordinate 0 and is used to mark the skyline's termination where the rightmost building ends. Any ground between the leftmost and rightmost buildings should be part of the skyline's contour.

Note: There must be no consecutive horizontal lines of equal height in the output skyline. For instance, [...,[2 3],[4 5],[7 5],[11 5],[12 7],...] is not acceptable; the three lines of height 5 should be merged into one in the final output as such: [...,[2 3],[4 5],[12 7],...]

Input & Output

Example 1 — Overlapping Buildings

$ Input: buildings = [[2,9,10],[3,7,15],[5,12,12]]

› Output: [[2,10],[3,15],[7,12],[9,12],[12,0]]

💡 Note: At x=2: building 1 starts (height 10). At x=3: building 2 starts (height 15, new max). At x=7: building 2 ends (height drops to 12 from building 3). At x=9: building 1 ends (height stays 12). At x=12: building 3 ends (height drops to 0).

Example 2 — Adjacent Buildings

$ Input: buildings = [[0,2,3],[2,5,3]]

› Output: [[0,3],[5,0]]

💡 Note: Two adjacent buildings with same height merge into one continuous skyline segment from x=0 to x=5.

Example 3 — Single Building

$ Input: buildings = [[1,3,2]]

› Output: [[1,2],[3,0]]

💡 Note: Simple case: building starts at x=1 with height 2, ends at x=3 returning to ground level 0.

Constraints

1 ≤ buildings.length ≤ 10⁴
0 ≤ left_i < right_i ≤ 2³¹ - 1
1 ≤ height_i ≤ 2³¹ - 1
buildings is sorted by left_i in non-decreasing order

Visualization

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Understanding the Visualization

Input Buildings

Buildings as [left, right, height] rectangles

Find Key Points

Identify where skyline height changes

Output Skyline

List of [x, height] key points

Key Takeaway

🎯 Key Insight: The skyline only changes at building start/end points, so we can use a sweep line algorithm to efficiently process these critical events.

Asked in

G Google 45 f Facebook 38 a Amazon 32 M Microsoft 28 U Uber 22

The key insight is to treat this as a sweep line problem where we only care about events when buildings start or end. The best approach uses a sweep line with priority queue to efficiently track the maximum height at each critical point. Time: O(n log n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Coordinate Scanning	O(n³)	O(n)	Check every x-coordinate and find the maximum height at each point
Sweep Line with Priority Queue	O(n log n)	O(n)	Use sweep line algorithm with max heap to efficiently track height changes

Brute Force Coordinate Scanning — Algorithm Steps

Collect all unique x-coordinates from building edges
For each x-coordinate, scan all buildings to find max height
Record points where height changes from previous

Visualization

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Step-by-Step Walkthrough

Collect Coordinates

Gather all building start/end points

Scan Each Point

For each x, check all buildings for max height

Record Changes

Add key points when height changes

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int** solution(int** buildings, int buildingsSize, int* buildingsColSize, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (buildingsSize == 0) return NULL;
    
    // Collect all unique x-coordinates
    int coords[2000];
    int coordCount = 0;
    
    for (int i = 0; i < buildingsSize; i++) {
        // Add left edge
        int found = 0;
        for (int j = 0; j < coordCount; j++) {
            if (coords[j] == buildings[i][0]) {
                found = 1;
                break;
            }
        }
        if (!found) coords[coordCount++] = buildings[i][0];
        
        // Add right edge
        found = 0;
        for (int j = 0; j < coordCount; j++) {
            if (coords[j] == buildings[i][1]) {
                found = 1;
                break;
            }
        }
        if (!found) coords[coordCount++] = buildings[i][1];
    }
    
    // Sort coordinates
    for (int i = 0; i < coordCount - 1; i++) {
        for (int j = i + 1; j < coordCount; j++) {
            if (coords[i] > coords[j]) {
                int temp = coords[i];
                coords[i] = coords[j];
                coords[j] = temp;
            }
        }
    }
    
    int** result = (int**)malloc(coordCount * sizeof(int*));
    *returnColumnSizes = (int*)malloc(coordCount * sizeof(int));
    int prevHeight = 0;
    
    for (int i = 0; i < coordCount; i++) {
        int x = coords[i];
        
        // Find max height at coordinate x
        int maxHeight = 0;
        for (int j = 0; j < buildingsSize; j++) {
            int left = buildings[j][0], right = buildings[j][1], height = buildings[j][2];
            if (left <= x && x < right) {
                if (height > maxHeight) {
                    maxHeight = height;
                }
            }
        }
        
        // Add key point if height changed
        if (maxHeight != prevHeight) {
            result[*returnSize] = (int*)malloc(2 * sizeof(int));
            result[*returnSize][0] = x;
            result[*returnSize][1] = maxHeight;
            (*returnColumnSizes)[*returnSize] = 2;
            (*returnSize)++;
            prevHeight = maxHeight;
        }
    }
    
    return result;
}

int main() {
    // Simple parsing for demo - in practice would need more robust JSON parsing
    int buildings[100][3];
    int buildingsSize = 3;
    
    // Example: [[2,9,10],[3,7,15],[5,12,12]]
    buildings[0][0] = 2; buildings[0][1] = 9; buildings[0][2] = 10;
    buildings[1][0] = 3; buildings[1][1] = 7; buildings[1][2] = 15;
    buildings[2][0] = 5; buildings[2][1] = 12; buildings[2][2] = 12;
    
    int* buildingsColSize = (int*)malloc(buildingsSize * sizeof(int));
    for (int i = 0; i < buildingsSize; i++) buildingsColSize[i] = 3;
    
    int** buildingsPtr = (int**)malloc(buildingsSize * sizeof(int*));
    for (int i = 0; i < buildingsSize; i++) buildingsPtr[i] = buildings[i];
    
    int returnSize;
    int* returnColumnSizes;
    int** result = solution(buildingsPtr, buildingsSize, buildingsColSize, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("[%d,%d]", result[i][0], result[i][1]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of O(n) coordinates, we scan O(n) buildings, and there can be O(n) coordinates

⚠ Quadratic Growth

Space Complexity

O(n)

Storage for coordinate set and result array

⚡ Linearithmic Space

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Ln 1, Col 1

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Coordinate Scanning — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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