Valid Boomerang - Practice Coding Problems

Valid Boomerang - Problem

Given an array points where points[i] = [xi, yi] represents a point on the X-Y plane, determine if these three points form a valid boomerang.

A boomerang is a set of three points that are:

All distinct (no two points are the same)
Not collinear (not in a straight line)

Think of it as checking if three points can form a triangle - if they're all on the same line, you can't make a triangle (or a boomerang shape)!

Goal: Return true if the three points form a valid boomerang, false otherwise.

Example: Points [[1,1], [2,3], [3,2]] form a boomerang because they create a triangle, while [[1,1], [2,2], [3,3]] don't because they're all on the same diagonal line.

Input & Output

example_1.py — Basic Valid Boomerang

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

› Output: true

💡 Note: These three points form a triangle (not collinear), so they make a valid boomerang. The cross product calculation: (2-1)*(2-1) - (3-1)*(3-1) = 1*1 - 2*2 = -3 ≠ 0

example_2.py — Collinear Points

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

› Output: false

💡 Note: All three points lie on the same diagonal line y=x, so they cannot form a boomerang. The cross product: (2-1)*(3-1) - (2-1)*(3-1) = 1*2 - 1*2 = 0

example_3.py — Duplicate Points

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

› Output: false

💡 Note: The first and third points are identical, so we don't have three distinct points. The cross product: (2-1)*(1-1) - (2-1)*(1-1) = 1*0 - 1*0 = 0

Constraints

points.length == 3
points[i].length == 2
-100 ≤ x_i, y_i ≤ 100
The input will always contain exactly three points

Visualization

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Understanding the Visualization

Position the Posts

Place three fence posts at the given coordinates

Form Vectors

Create directional arrows from the first post to the other two

Calculate Cross Product

Use the cross product formula to measure the 'turning angle'

Check Result

If cross product is zero, posts are aligned (straight fence). If non-zero, they form a triangle (valid boomerang)!

Key Takeaway

🎯 Key Insight: The cross product method elegantly determines collinearity without division, making it the most robust solution for checking if three points can form a boomerang!

Asked in

G Google 15 a Amazon 8 f Meta 12 ⊞ Microsoft 6

The cross product method is optimal for this problem. Calculate the cross product of vectors formed by the three points: (x₂-x₁)(y₃-y₁) - (y₂-y₁)(x₃-x₁). If the result is zero, the points are collinear (not a boomerang). This approach avoids division operations and floating-point precision issues, making it more robust than slope comparison.

Common Approaches

Approach	Time	Space	Notes
✓ Cross Product Method (Optimal)	O(1)	O(1)	Use cross product to check if three points are collinear without division

Cross Product Method (Optimal) — Algorithm Steps

Extract the three points from the input array
Create two vectors: from point 1 to point 2, and from point 1 to point 3
Calculate the cross product of these vectors
If cross product is zero, points are collinear (not a boomerang)
If cross product is non-zero, points form a valid boomerang

Visualization

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Step-by-Step Walkthrough

Create Vectors

Form vectors from point 1 to points 2 and 3

Calculate Cross Product

Compute (x₂-x₁)(y₃-y₁) - (y₂-y₁)(x₃-x₁)

Check Result

Non-zero cross product means valid boomerang

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool isBoomerang(int points[][2]) {
    // Extract points
    int x1 = points[0][0], y1 = points[0][1];
    int x2 = points[1][0], y2 = points[1][1];
    int x3 = points[2][0], y3 = points[2][1];
    
    // Create vectors from point 1 to point 2 and point 1 to point 3
    // Vector 1: (x2-x1, y2-y1)
    // Vector 2: (x3-x1, y3-y1)
    
    // Calculate cross product: (x2-x1)*(y3-y1) - (y2-y1)*(x3-x1)
    // If cross product is 0, points are collinear
    int crossProduct = (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
    
    return crossProduct != 0;
}

int main() {
    int points[3][2];
    for (int i = 0; i < 3; i++) {
        scanf("%d %d", &points[i][0], &points[i][1]);
    }
    printf("%s\n", isBoomerang(points) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of operations regardless of input (always 3 points)

✓ Linear Growth

Space Complexity

O(1)

Only using variables for coordinate differences

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Cross Product Method (Optimal) — Algorithm Steps

Visualization

Code -

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