Solution without calculating any distances, 4 lines (Python)


  • 0
    D

    Simply sort the points and figure out where the last two points should be with the coordinates of the first two. There is no need to calculate any distances.

    def validSquare(self, p1, p2, p3, p4):
        if p1 == p2 == p3 == p4: return False
    
        p1,p2,p3,p4 = sorted([p1,p2,p3,p4])
        if p2[1] < p3[1]: p2,p3 = p3,p2
    
        return p3 == [p1[0] + (p2[1]-p1[1]), p1[1] - (p2[0]-p1[0])]\
           and p4 == [p2[0] + (p2[1]-p1[1]), p2[1] - (p2[0]-p1[0])]

Log in to reply
 

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.