Rectangles Area - Practice Coding Problems

Rectangles Area - Problem

Database Medium

Imagine you have a collection of points scattered across a coordinate plane, and you want to find all possible axis-aligned rectangles that can be formed using any two of these points as opposite corners!

Given a table Points with columns:

id - unique identifier for each point
x_value - x-coordinate of the point
y_value - y-coordinate of the point

Your mission: Find all possible rectangles with non-zero area that can be formed by selecting any two points as opposite corners. Each rectangle must be axis-aligned (sides parallel to x and y axes).

Return format: Each row should contain (p1, p2, area) where:

p1 and p2 are the IDs of the two corner points
area is the calculated area (must be > 0)

Sorting: Order by area (descending), then by p1 (ascending), then by p2 (ascending).

Input & Output

example_1.sql — Basic Rectangle Formation

$ Input: Points table: +----+---------+---------+ | id | x_value | y_value | +----+---------+---------+ | 1 | 1 | 1 | | 2 | 1 | 3 | | 3 | 3 | 1 | | 4 | 3 | 3 | +----+---------+---------+

› Output: +----+----+------+ | p1 | p2 | area | +----+----+------+ | 1 | 4 | 4 | | 2 | 3 | 4 | +----+----+------+

💡 Note: Points (1,1) and (3,3) form a 2×2 rectangle with area 4. Points (1,3) and (3,1) also form a 2×2 rectangle with area 4. Both have same area, so ordered by p1.

example_2.sql — Different Sized Rectangles

$ Input: Points table: +----+---------+---------+ | id | x_value | y_value | +----+---------+---------+ | 1 | 0 | 0 | | 2 | 4 | 0 | | 3 | 0 | 3 | | 4 | 4 | 4 | +----+---------+---------+

› Output: +----+----+------+ | p1 | p2 | area | +----+----+------+ | 1 | 4 | 16 | | 2 | 3 | 12 | +----+----+------+

💡 Note: Point (0,0) and (4,4) create a 4×4 rectangle (area=16). Point (4,0) and (0,3) create a 4×3 rectangle (area=12). Ordered by area descending.

example_3.sql — No Valid Rectangles

$ Input: Points table: +----+---------+---------+ | id | x_value | y_value | +----+---------+---------+ | 1 | 1 | 1 | | 2 | 1 | 2 | | 3 | 1 | 3 | +----+---------+---------+

› Output: +----+----+------+ | p1 | p2 | area | +----+----+------+ (empty result)

💡 Note: All points have the same x-value (1), so no valid axis-aligned rectangles can be formed. Need points with different x AND y coordinates.

Constraints

2 ≤ number of points ≤ 500
0 ≤ x_value, y_value ≤ 10⁶
All point coordinates are integers
id values are unique for each point
Points may have duplicate coordinates (same x_value and y_value)

Visualization

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Understanding the Visualization

Identify Valid Pairs

Two points can form a rectangle only if they have different x-coordinates AND different y-coordinates

Visualize Rectangle

The two points become opposite corners of an axis-aligned rectangle

Calculate Area

Area equals the absolute difference in x-values times the absolute difference in y-values

Sort and Return

Order results by area (descending), then by first point ID, then by second point ID

Key Takeaway

🎯 Key Insight: Use SQL self-join to examine all point pairs, filter for different x AND y coordinates, calculate rectangular areas, and sort results appropriately.

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Meta 18

This database problem requires finding all possible axis-aligned rectangles from point pairs. The key insight is that two points can form a rectangle if they have different x AND y coordinates. Use a self-join to examine all point pairs, filter for valid rectangles, calculate areas using |x2-x1| × |y2-y1|, and sort appropriately.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force SQL Query	O(n²)	O(n²)	Check all pairs of points using cross join and calculate areas

Brute Force SQL Query — Algorithm Steps

Perform self-join on Points table to get all point pairs
Filter pairs where points have different x_value AND y_value
Ensure p1.id < p2.id to avoid duplicate pairs
Calculate area as ABS(x2-x1) × ABS(y2-y1)
Filter out zero areas
Sort by area DESC, p1 ASC, p2 ASC

Visualization

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Step-by-Step Walkthrough

Select Point Pairs

Choose two points with different x AND y coordinates

Form Rectangle

Points become opposite corners of axis-aligned rectangle

Calculate Area

Area = |x2-x1| × |y2-y1|

Sort Results

Order by area desc, then p1 asc, then p2 asc

Code -

solution.c — C

// SQL Query
/*
SELECT 
    p1.id as p1,
    p2.id as p2,
    ABS(p2.x_value - p1.x_value) * ABS(p2.y_value - p1.y_value) as area
FROM Points p1
CROSS JOIN Points p2
WHERE p1.id < p2.id
    AND p1.x_value != p2.x_value
    AND p1.y_value != p2.y_value
ORDER BY area DESC, p1 ASC, p2 ASC;
*/

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

typedef struct {
    int id, x_value, y_value;
} Point;

typedef struct {
    int p1, p2, area;
} Rectangle;

int abs_val(int x) {
    return x < 0 ? -x : x;
}

int compare(const void* a, const void* b) {
    Rectangle* ra = (Rectangle*)a;
    Rectangle* rb = (Rectangle*)b;
    if (ra->area != rb->area) return rb->area - ra->area;
    if (ra->p1 != rb->p1) return ra->p1 - rb->p1;
    return ra->p2 - rb->p2;
}

int solution(Point* points, int n, Rectangle* result) {
    int count = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            Point pt1 = points[i], pt2 = points[j];
            if (pt1.x_value != pt2.x_value && pt1.y_value != pt2.y_value) {
                int area = abs_val(pt2.x_value - pt1.x_value) * abs_val(pt2.y_value - pt1.y_value);
                if (area > 0) {
                    result[count].p1 = pt1.id;
                    result[count].p2 = pt2.id;
                    result[count].area = area;
                    count++;
                }
            }
        }
    }
    
    qsort(result, count, sizeof(Rectangle), compare);
    return count;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Must examine every pair of points in the table

⚠ Quadratic Growth

Space Complexity

O(n²)

Result set can contain up to n² rectangles in worst case

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force SQL Query — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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