Valid Boomerang - Practice Coding Problems

Valid Boomerang - Problem

Given an array points where points[i] = [x_i, y_i] represents a point on the X-Y plane, return true if these points form a boomerang.

A boomerang is a set of three points that are all distinct and not in a straight line.

Input & Output

Example 1 — Valid Boomerang

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

› Output: true

💡 Note: The three points are distinct and not collinear, so they form a valid boomerang. Cross product calculation: (2-1)×(2-1) - (3-1)×(3-1) = 1×1 - 2×2 = -3 ≠ 0

Example 2 — Collinear Points

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

› Output: false

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

Example 3 — Duplicate Points

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

› Output: false

💡 Note: Two points are identical ([1,1] appears twice), so they cannot form a valid boomerang. Cross product: (2-1)×(1-1) - (3-1)×(1-1) = 1×0 - 2×0 = 0

Constraints

points.length == 3
points[i].length == 2
0 ≤ points[i][j] ≤ 50

Visualization

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Understanding the Visualization

Input

Three points on 2D plane: [[1,1],[2,3],[3,2]]

Process

Check if points are distinct and not on same line

Output

Return true if valid boomerang, false otherwise

Key Takeaway

🎯 Key Insight: Cross product elegantly detects collinearity without division edge cases

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use the cross product of vectors to check if three points are collinear. Three points form a valid boomerang if they are distinct and not collinear. Best approach is the cross product method which avoids division and handles all edge cases naturally. Time: O(1), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Slope Comparison	O(1)	O(1)	Calculate slopes between point pairs and check if they're different
Cross Product Method	O(1)	O(1)	Use cross product of vectors to check collinearity without division

Slope Comparison — Algorithm Steps

Check if any two points are identical
Calculate slope between points 0-1 and 0-2
Compare slopes - if different, return true

Visualization

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Step-by-Step Walkthrough

Check Points

Verify no two points are identical

Calculate Slopes

Compare slopes between point pairs

Result

Different slopes mean valid boomerang

Code -

solution.c — C

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

int solution(int points[3][2]) {
    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];
    
    // Check if any two points are the same
    if ((x1 == x2 && y1 == y2) || (x1 == x3 && y1 == y3) || (x2 == x3 && y2 == y3)) {
        return 0;
    }
    
    // Check if points are collinear using cross multiplication
    return (y2 - y1) * (x3 - x1) != (y3 - y1) * (x2 - x1);
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int points[3][2];
    
    // Simple parsing for [[x1,y1],[x2,y2],[x3,y3]] format
    char *ptr = line;
    int idx = 0;
    
    while (*ptr && idx < 3) {
        if (*ptr == '[') {
            ptr++;
            if (idx > 0) { // Skip outer bracket for first point
                sscanf(ptr, "%d,%d", &points[idx][0], &points[idx][1]);
                idx++;
            } else {
                ptr++; // Skip second bracket
                sscanf(ptr, "%d,%d", &points[idx][0], &points[idx][1]);
                idx++;
            }
        }
        ptr++;
    }
    
    int result = solution(points);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only performs constant number of operations for three points

✓ Linear Growth

Space Complexity

O(1)

Uses only a few variables for slope calculations

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Slope Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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