Valid Boomerang — Easy Problem & Solution

Problem statement

points holds exactly three points [x, y] in the plane. They form a boomerang when all three are different and they do not lie on one straight line.

Return true if the points form a boomerang.

Example 1

Input: points = [[1,1],[2,3],[3,2]]
Output: true

Example 2

Input: points = [[1,1],[2,2],[3,3]]
Output: false
Explanation: All three lie on the line y = x.

Example 3

Input: points = [[0,0],[0,0],[4,7]]
Output: false
Explanation: Two of the points coincide.

Constraints

• points.length == 3
• points[i].length == 2
• 0 <= xi, yi <= 100

How to solve Valid Boomerang

The cross product of p2 - p1 and p3 - p1 is twice the signed area of the triangle. It is zero exactly when the three points are collinear — which includes the case where two of them coincide.

Approach

1. Let (ax, ay) = p2 - p1 and (bx, by) = p3 - p1.
2. Compute ax * by - ay * bx.
3. Return true if it is non-zero.

Why it works

The cross product of two vectors is zero exactly when one is a scalar multiple of the other (or either is zero). If p3 - p1 is a multiple of p2 - p1, then p3 lies on the line through p1 and p2; if either vector is zero, two points coincide. A non-zero cross product rules out both, which is exactly the boomerang condition. Everything stays in integers, so there is no slope division and no rounding.

Complexity

• Time — O(1)
• Space — O(1)

Pitfalls

• Comparing slopes with division breaks on vertical lines and on floating-point rounding; cross-multiply instead.
• Do not forget duplicates — the cross product already handles them, a separate equality check is redundant but harmless.

Reference solution

Python

from typing import List

def isBoomerang(points: List[List[int]]) -> bool:
    ax = points[1][0] - points[0][0]
    ay = points[1][1] - points[0][1]
    bx = points[2][0] - points[0][0]
    by = points[2][1] - points[0][1]
    return ax * by - ay * bx != 0

JavaScript

var isBoomerang = function(points) {
    var ax = points[1][0] - points[0][0];
    var ay = points[1][1] - points[0][1];
    var bx = points[2][0] - points[0][0];
    var by = points[2][1] - points[0][1];
    return ax * by - ay * bx !== 0;
};

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