Shortest Distance in a Plane

Shortest Distance in a Plane - Problem

Database Medium

You have a table Point2D that stores coordinates of points on a 2D plane. Each row represents a point with its (x, y) coordinates.

Your task: Find the shortest distance between any two points in the table.

The distance between two points p1(x1, y1) and p2(x2, y2) is calculated using the Euclidean distance formula: sqrt((x2 - x1)² + (y2 - y1)²)

Requirements:

Round the result to 2 decimal places
Return only the shortest distance value
Handle cases where multiple point pairs have the same minimum distance

Table Schema

Point2D

Column Name	Type	Description
`x` PK	int	X coordinate of the point
`y` PK	int	Y coordinate of the point

Primary Key: (x, y)

Note: The combination of (x, y) is unique for each point

Input & Output

Example 1 — Basic Point Distance

Input Table:

x	y
-1	-1
0	0
-1	1

Output:

shortest
1.41

💡 Note:

We have three points: (-1,-1), (0,0), and (-1,1). The distances are:

(-1,-1) to (0,0): √[(0-(-1))² + (0-(-1))²] = √2 ≈ 1.41
(-1,-1) to (-1,1): √[(-1-(-1))² + (1-(-1))²] = √4 = 2.00
(0,0) to (-1,1): √[(-1-0)² + (1-0)²] = √2 ≈ 1.41

The minimum distance is 1.41.

Example 2 — Same Distance Points

Input Table:

x	y
0	0
3	4

Output:

shortest
5

💡 Note:

With only two points (0,0) and (3,4), there's only one distance to calculate:

Distance = √[(3-0)² + (4-0)²] = √[9 + 16] = √25 = 5.00

Example 3 — Close Points

Input Table:

x	y
0	0
1	0
0	1
2	2

Output:

shortest
1

💡 Note:

Multiple pairs have distance 1.00: (0,0) to (1,0) and (0,0) to (0,1). Both are horizontal/vertical distances of 1 unit, which is the minimum among all possible pairs.

Constraints

2 ≤ number of points ≤ 500
-10^4 ≤ x, y ≤ 10^4
All points have distinct coordinates
At least one pair of points exists

Visualization

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Asked in

A Amazon 15 G Google 12 M Microsoft 8

Use self-join to create point pairs, apply SQRT(POWER(x2-x1,2) + POWER(y2-y1,2)) for Euclidean distance, and find MIN() distance rounded to 2 decimal places.

Table Schema

Point2D

Column Name	Type	Description
`x` PK	int	X coordinate of the point
`y` PK	int	Y coordinate of the point

Primary Key: (x, y)

Note: The combination of (x, y) is unique for each point

Common Approaches

✓ Self-Join with Distance Calculation

⏱️ Time: O(n²) Space: O(n²)

Use self-join to generate all unique point pairs, apply the distance formula, and find the minimum distance. The key is to avoid comparing a point with itself and prevent duplicate pairs.

Self-Join with Distance Calculation — Algorithm Steps

Step 1: Self-join Point2D to create all point combinations
Step 2: Filter to avoid same points and duplicate pairs (p1 < p2)
Step 3: Calculate Euclidean distance using SQRT and POWER functions
Step 4: Find minimum distance and round to 2 decimal places

Visualization

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Step-by-Step Walkthrough

Self-Join

Create all point pairs

Filter

Avoid duplicates with p1 < p2

Calculate

Apply distance formula

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Must compare every pair of points

⚠ Quadratic Growth

Space Complexity

O(n²)

Temporary storage for all point pairs

⚠ Quadratic Space

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Table Schema

Input & Output

Constraints

Visualization

Related Problems

Table Schema

Common Approaches

Self-Join with Distance Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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