forked from shuboc/LeetCode-2
-
Notifications
You must be signed in to change notification settings - Fork 0
/
number-of-boomerangs.py
58 lines (51 loc) · 1.65 KB
/
number-of-boomerangs.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
# Time: O(n^2)
# Space: O(n)
# 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]]
import collections
class Solution(object):
def numberOfBoomerangs(self, points):
"""
:type points: List[List[int]]
:rtype: int
"""
result = 0
for i in xrange(len(points)):
group = collections.defaultdict(int)
for j in xrange(len(points)):
if j == i:
continue
dx, dy = points[i][0] - points[j][0], points[i][1] - points[j][1]
group[dx**2 + dy**2] += 1
for _, v in group.iteritems():
if v > 1:
result += v * (v-1)
return result
def numberOfBoomerangs2(self, points):
"""
:type points: List[List[int]]
:rtype: int
"""
cnt = 0
for a, i in enumerate(points):
dis_list = []
for b, k in enumerate(points[:a] + points[a + 1:]):
dis_list.append((k[0] - i[0]) ** 2 + (k[1] - i[1]) ** 2)
for z in collections.Counter(dis_list).values():
if z > 1:
cnt += z * (z - 1)
return cnt