LeetCode 149. Max Points on a Line

Problem Statement

(Source) Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Solution

Tags:

Hash Table

,

Math

.

Note: In Java version of pre-defined code, coordinates of points are pre-defined as integers. So I assume the same applies here in python version.

class Solution(object):
def maxPoints(self, points):
"""
:type points: List[Point]
:rtype: int
"""
res = 0
n = len(points)
for i in xrange(n):
memo = {}
duplicate = 0
for j in xrange(n):
if (points[j].x, points[j].y) == (points[i].x, points[i].y):
duplicate += 1
elif points[j].x == points[i].x:
memo['v'] = memo.get('v', 0) + 1
elif points[j].y == points[i].y:
memo['h'] = memo.get('h', 0) + 1
else:
k = self.slope(points[i], points[j])
memo[k] = memo.get(k, 0) + 1
res = max(res, max(memo.values()) + duplicate if memo else duplicate)
return res

def gcd(self, x, y):
while x % y: x, y = y, x%y
return y

def slope(self, p1, p2):
res = ''
flag = 1
x = p1.x - p2.x
y = p1.y - p2.y
if x < 0:
flag *= -1
x = -x
if y < 0:
flag *= -1
y = -y
m = self.gcd(x, y)
if flag == -1:
res='-'
res += (str(x/m) + '/' + str(y/m))
return res

Complexity analysis:

Time complexity: O(n2⋅gcd). For time complexity analysis of

gcd

, see below reference.

References

(1) Time complexity of Euclid’s Algorithm.