# Solution works but... Brute force, Time exceeded

• ``````class Solution(object):
def numberOfBoomerangs(self,points):
"""Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k)
such that the distance between i and j equals the distance between i and k (the order of the tuple matters).

Find the number of boomerangs. You may assume that n will be at most 500 and
coordinates of points are all in the range [-10000, 10000] (inclusive).

Example:
Input:
[[0,0],[1,0],[2,0]]

Output:
2

Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]
"""

# points.sort()

boomerangs = []
for i in range(len(points)):
counter1 =i+1
while counter1 <= len(points)-2:
counter2=counter1+1
while counter2 <= len(points) - 1:
a = points[i]
b = points[counter1]
c = points[counter2]
self.isBoomarang(a, b, c, boomerangs)
counter2 += 1
counter1+=1
return len(boomerangs)

def isBoomarang(self, a, b, c, boomerangs):
# print a, b, c
pointABdistance =(a[0]-b[0])**2 + (a[1]-b[1])**2
pointACdistance =(a[0]-c[0])**2 + (a[1]-c[1])**2
pointBCdistance = (b[0] - c[0]) ** 2 + (b[1] - c[1]) ** 2

if pointABdistance == pointACdistance:
boomerangs.append([a, b, c])
boomerangs.append([a, c, b ])
if pointABdistance == pointBCdistance:
boomerangs.append([b, a, c])
boomerangs.append([b, c, a])
if pointACdistance == pointBCdistance:
boomerangs.append([c, b, a,])
boomerangs.append([c, a, b])
return (pointABdistance == pointACdistance) or  (pointBCdistance == pointACdistance) or pointABdistance == pointBCdistance
``````

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