### View *Valid Boomerang* on LeetCode

## Statistics

**Time Complexity**

O(n) - Each pair in `points`

must be visited, resulting in the O(n) time complexity.

**Space Complexity**

O(n) - Each pair in `points`

must be stored, resulting in the O(n) space complexity. However, it could be argued that it is O(1) because the input is a constant number of elements (three pairs of two).

**Runtime Beats**

64.29% of other submissions

**Memory Beats**

67.95% of other sumbissions

## Explanation

The program implements the formula for calculating a triangle. If the input violates the rules, a triangle cannot be created, resulting in `area`

equaling zero.

## Solution

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class Solution:
def isBoomerang(self, points: List[List[int]]) -> bool:
x1, y1 = points[0]
x2, y2 = points[1]
x3, y3 = points[2]
area = abs(x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2))/2
return area != 0