Valid Boomerang

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

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