Widest Vertical Area Between Two Points Containing No Points

Widest Vertical Area Between Two Points Containing No Points - Problem

Array Sorting Easy

Imagine you're a city planner looking at a map with n buildings represented as points on a 2D plane, where each point points[i] = [xi, yi] has coordinates (x, y).

Your task is to find the widest vertical corridor between any two buildings such that no other buildings obstruct the path. Think of it as finding the widest north-south street you can build!

A vertical area is like an invisible corridor of fixed width that extends infinitely along the y-axis (infinite height). The widest vertical area is the one with maximum width between two x-coordinates.

Important: Buildings exactly on the edge of a corridor don't count as obstructions - they can be right next to the street!

Goal: Return the maximum width of any such vertical corridor.

Input & Output

example_1.py — Basic Case

$ Input: points = [[8,7],[9,9],[7,4],[9,7]]

› Output: 1

💡 Note: After sorting x-coordinates: [7,8,9,9]. The gaps are: 8-7=1, 9-8=1, 9-9=0. Maximum gap is 1.

example_2.py — Larger Gap

$ Input: points = [[3,1],[9,0],[1,0],[1,4],[5,3],[8,8]]

› Output: 3

💡 Note: After sorting x-coordinates: [1,1,3,5,8,9]. The gaps are: 1-1=0, 3-1=2, 5-3=2, 8-5=3, 9-8=1. Maximum gap is 3.

example_3.py — Minimum Case

$ Input: points = [[1,1],[2,2]]

› Output: 1

💡 Note: With only two points at x=1 and x=2, the only possible vertical area has width 2-1=1.

Visualization

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Understanding the Visualization

Ignore Building Heights

Y-coordinates don't matter for vertical streets - only east-west positions (x-coordinates) matter

Line Up Buildings

Sort all buildings by their east-west position from left to right

Measure Street Widths

Check the distance between each pair of adjacent buildings

Find Widest Street

The largest gap between adjacent buildings is your answer

Key Takeaway

🎯 Key Insight: Y-coordinates don't matter for vertical areas! Sort by x-coordinate and find the maximum gap between adjacent positions.

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of the n² pairs of points, we check all n points to see if they lie between

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum distance

✓ Linear Space

Constraints

n == points.length
2 ≤ n ≤ 10⁵
points[i].length == 2
0 ≤ xi, yi ≤ 10⁹

Asked in

a Amazon 15 G Google 12 ⊞ Microsoft 8 Apple 5

The optimal approach is to sort the points by x-coordinate and find the maximum gap between consecutive x-coordinates. This works because y-coordinates are irrelevant for vertical areas, and the widest area must occur between two adjacent x-coordinates in sorted order.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check All Pairs)	O(n³)	O(1)	Check every pair of points and verify no other points lie between them
Sorting + Gap Finding (Optimal)	O(n log n)	O(n)	Sort points by x-coordinate and find the maximum gap between consecutive points

Brute Force (Check All Pairs) — Algorithm Steps

For each pair of points (i, j)
Calculate horizontal distance |xi - xj|
Check if any other point k has xk between xi and xj
If no points in between, update maximum distance
Return the maximum distance found

Visualization

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Step-by-Step Walkthrough

Select First Pair

Choose points A and B, calculate horizontal distance

Check All Points

Verify no other points have x-coordinates between A and B

Update Maximum

If valid, update maximum width found

Repeat

Continue for all possible pairs

Code -

solution.c — C

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

int largestWidestArea(int points[][2], int n) {
    int maxWidth = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int x1 = points[i][0], x2 = points[j][0];
            if (x1 > x2) {
                int temp = x1; x1 = x2; x2 = temp;
            }
            
            // Check if any point lies strictly between x1 and x2
            int valid = 1;
            for (int k = 0; k < n; k++) {
                if (k != i && k != j) {
                    int xk = points[k][0];
                    if (x1 < xk && xk < x2) {
                        valid = 0;
                        break;
                    }
                }
            }
            
            if (valid) {
                int width = x2 - x1;
                if (width > maxWidth) {
                    maxWidth = width;
                }
            }
        }
    }
    
    return maxWidth;
}

int main() {
    int n;
    scanf("%d", &n);
    int points[n][2];
    for (int i = 0; i < n; i++) {
        scanf("%d %d", &points[i][0], &points[i][1]);
    }
    printf("%d\n", largestWidestArea(points, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of the n² pairs of points, we check all n points to see if they lie between

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum distance

✓ Linear Space

Constraints

n == points.length
2 ≤ n ≤ 10⁵
points[i].length == 2
0 ≤ xi, yi ≤ 10⁹

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Check All Pairs) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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