Shortest Distance in a Line - Practice Coding Problems

Shortest Distance in a Line - Problem

Database Easy

Find the Shortest Distance Between Any Two Points

You have a database table called Point that contains coordinates of points positioned on the X-axis. Your task is to find the shortest distance between any two points in this table.

Table Schema:

+-------------+------+
| Column Name | Type |
+-------------+------+
| x           | int  |
+-------------+------+

Where x is the primary key representing the position of a point on the X-axis.

Goal: Write a SQL query to return the minimum absolute difference between any two points.

Example: If you have points at positions [1, 3, 6, 10], the shortest distance would be between points 1 and 3, which gives us a distance of |3 - 1| = 2.

Input & Output

example_1.sql — Basic Case

$ Input: Point table: +---+ | x | +---+ | 1 | | 3 | | 6 | |10 | +---+

› Output: +-------------------+ | shortest_distance | +-------------------+ | 2 | +-------------------+

💡 Note: The distances between consecutive points are: |3-1|=2, |6-3|=3, |10-6|=4. The minimum is 2.

example_2.sql — Two Points Only

$ Input: Point table: +---+ | x | +---+ |-5 | | 7 | +---+

› Output: +-------------------+ | shortest_distance | +-------------------+ | 12 | +-------------------+

💡 Note: With only two points at -5 and 7, the distance is |7-(-5)| = 12.

example_3.sql — Negative Coordinates

$ Input: Point table: +----+ | x | +----+ |-10| | -2| | 0| | 5| +----+

› Output: +-------------------+ | shortest_distance | +-------------------+ | 2 | +-------------------+

💡 Note: Sorted points: [-10, -2, 0, 5]. Distances: |-2-(-10)|=8, |0-(-2)|=2, |5-0|=5. Minimum is 2.

Constraints

2 ≤ number of points ≤ 10⁶
-10⁸ ≤ x ≤ 10⁸
All points have unique x-coordinates (primary key constraint)
Result will always fit in a 32-bit signed integer

Visualization

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Understanding the Visualization

Plot points on number line

Visualize all points as positions on a straight line

Sort by position

Arrange points from left to right (smallest to largest x-coordinate)

Check adjacent distances

The closest pair must be neighbors when sorted

Return minimum

The smallest distance among all adjacent pairs is our answer

Key Takeaway

🎯 Key Insight: Once points are sorted by position, the shortest distance must be between consecutive points, reducing complexity from O(n²) to O(n log n).

Asked in

G Google 45 a Amazon 38 ⊞ Microsoft 32 f Meta 25

The optimal solution uses SQL window functions with LEAD() to compare only adjacent points after sorting. Key insight: the minimum distance between any two points on a line must occur between consecutive points when sorted by position. Time complexity: O(n log n) for sorting, much better than the O(n²) brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force SQL (Cartesian Product)	O(n²)	O(n²)	Compare every point with every other point using self-join
Optimized SQL with Window Functions	O(n log n)	O(n)	Sort points and compare only adjacent pairs using window functions

Brute Force SQL (Cartesian Product) — Algorithm Steps

Self-join the Point table to create all combinations
Filter out pairs where both points are the same (p1.x < p2.x)
Calculate absolute difference for each pair
Find the minimum difference using MIN function

Visualization

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Step-by-Step Walkthrough

Create all pairs

Self-join creates all possible combinations of two different points

Calculate distances

Compute absolute difference for each pair

Find minimum

Return the smallest distance found

Code -

solution.c — C

// SQL Query for Brute Force Approach
char* sqlQuery = ""
"SELECT MIN(ABS(p1.x - p2.x)) AS shortest_distance
"
"FROM Point p1
"
"JOIN Point p2 ON p1.x < p2.x;";

// Example implementation if this were a C problem
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int shortestDistanceBruteForce(int points[], int n) {
    if (n < 2) return 0;
    
    int minDistance = INT_MAX;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int distance = abs(points[i] - points[j]);
            if (distance < minDistance) {
                minDistance = distance;
            }
        }
    }
    
    return minDistance;
}

int main() {
    int points[] = {1, 3, 6, 10};
    int n = sizeof(points) / sizeof(points[0]);
    printf("%d\n", shortestDistanceBruteForce(points, n)); // Output: 2
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Self-join creates n² pairs, then we scan all pairs to find minimum

⚠ Quadratic Growth

Space Complexity

O(n²)

Temporary result set stores all possible pairs of points

⚠ Quadratic Space

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Find the Shortest Distance Between Any Two Points

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force SQL (Cartesian Product) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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