Medium
Number Of Corner Rectangles — Python
Full explanation · Time O(n * m^2) · Space O(n * m)
# Time: O(n * m^2), n is the number of rows with 1s, m is the number of cols with 1s
# Space: O(n * m)
class Solution(object):
def countCornerRectangles(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
rows = [[c for c, val in enumerate(row) if val]
for row in grid]
result = 0
for i in xrange(len(rows)):
lookup = set(rows[i])
for j in xrange(i):
count = sum(1 for c in rows[j] if c in lookup)
result += count*(count-1)/2
return result