Max Points on a Line with python 3 - python-3.x

Question

algorithm question:
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
Example 1:
Input: [[1,1],[2,2],[3,3]]
Output: 3
Explanation:
^
|
| o
| o
| o
+------------->
0 1 2 3 4
Example 2:
Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]]
Output: 4
the working python 3 code is below:
wondering
snippet 1 d[slope] = d.get(slope, 1) + 1 is working
but why this snippet 2 is not working correctly for example 2 even though snippet 1 and 2 are the same
if slope in d:
d[slope] += 1
else:
d[slope] = 1
def gcd(self, a, b):
if b == 0:
return a
return self.gcd(b, a%b)
def get_slope(self, p1, p2):
dx = p1[0] - p2[0]
dy = p1[1] - p2[1]
c = self.gcd(dx, dy)
dx /= c
dy /= c
return str(dy) + "/" + str(dx)
def is_same_points(self, p1:List[int], p2:List[int]):
return p1[0] == p2[0] and p1[1] == p2[1]
def maxPoints(self, points: List[List[int]]) -> int:
if not points:
return 0
n = len(points)
count = 1
for i in range(0, n):
d = {}
duped = 0
localmax = 1
p1 = points[i]
for j in range(i+1, n):
p2 = points[j]
if self.is_same_points(p1, p2):
duped += 1
else:
slope = self.get_slope(p1, p2)
# 1) not work: output is 3 in example 2
# if slope in d:
# d[slope] += 1
# else:
# d[slope] = 1
# 2) works: correct output 4 for example 2
d[slope] = d.get(slope, 1) + 1
localmax = max(localmax, d[slope]);
count = max(count, localmax + duped)
return count

Answer 1

Interesting problem and nice solution.
The reason why the commented out code doesn't work is because of that:
else:
d[slope] = 1 ## correct would be d[slope] = 2
Every 2 points are on the same line, you are counting only one point for the first two p1 p2, thus you get one less in the final answer.