Medium
Number of Pairs of Interchangeable Rectangles — Python
Full explanation · Time O(n) · Space O(n)
# Time: O(n)
# Space: O(n)
import collections
import fractions
class Solution(object):
def interchangeableRectangles(self, rectangles):
"""
:type rectangles: List[List[int]]
:rtype: int
"""
count = collections.defaultdict(int)
for w, h in rectangles:
g = fractions.gcd(w, h) # Time: O(logx) ~= O(1)
count[(w//g, h//g)] += 1
return sum(c*(c-1)//2 for c in count.itervalues())