Count maximum points on same line

Given N point on a 2D plane as pair of (x, y) co-ordinates, we need to find maximum number of point which lie on the same line.
Examples: 

Input : points[] = {-1, 1}, {0, 0}, {1, 1}, 
                    {2, 2}, {3, 3}, {3, 4} 
Output : 4
Then maximum number of point which lie on same
line are 4, those point are {0, 0}, {1, 1}, {2, 2},
{3, 3}

We can solve above problem by following approach – For each point p, calculate its slope with other points and use a map to record how many points have same slope, by which we can find out how many points are on same line with p as their one point. For each point keep doing the same thing and update the maximum number of point count found so far.
Some things to note in implementation are: 
1) if two point are (x1, y1) and (x2, y2) then their slope will be (y2 – y1) / (x2 – x1) which can be a double value and can cause precision problems. To get rid of the precision problems, we treat slope as pair ((y2 – y1), (x2 – x1)) instead of ratio and reduce pair by their gcd before inserting into map. In below code points which are vertical or repeated are treated separately.
2) If we use unordered_map in c++ or HashMap in Java for storing the slope pair, then total time complexity of solution will be O(n^2)

C++

/* C/C++ program to find maximum number of point
which lie on same line */
#include <bits/stdc++.h>
#include <boost/functional/hash.hpp>

using namespace std;

// method to find maximum collinear point
int maxPointOnSameLine(vector< pair<int, int> > points)
{
    int N = points.size();
    if (N < 2)
        return N;

    int maxPoint = 0;
    int curMax, overlapPoints, verticalPoints;

    // here since we are using unordered_map
    // which is based on hash function
    //But by default we don't have hash function for pairs
    //so we'll use hash function defined in Boost library
    unordered_map<pair<int, int>, int,boost::
              hash<pair<int, int> > > slopeMap;

    // looping for each point
    for (int i = 0; i < N; i++)
    {
        curMax = overlapPoints = verticalPoints = 0;

        // looping from i + 1 to ignore same pair again
        for (int j = i + 1; j < N; j++)
        {
            // If both point are equal then just
            // increase overlapPoint count
            if (points[i] == points[j])
                overlapPoints++;

            // If x co-ordinate is same, then both
            // point are vertical to each other
            else if (points[i].first == points[j].first)
                verticalPoints++;

            else
            {
                int yDif = points[j].second - points[i].second;
                int xDif = points[j].first - points[i].first;
                int g = __gcd(xDif, yDif);

                // reducing the difference by their gcd
                yDif /= g;
                xDif /= g;

                // increasing the frequency of current slope
                // in map
                slopeMap[make_pair(yDif, xDif)]++;
                curMax = max(curMax, slopeMap[make_pair(yDif, xDif)]);
            }

            curMax = max(curMax, verticalPoints);
        }

        // updating global maximum by current point's maximum
        maxPoint = max(maxPoint, curMax + overlapPoints + 1);

        // printf("maximum collinear point
        // which contains current point
        // are : %d\n", curMax + overlapPoints + 1);
        slopeMap.clear();
    }

    return maxPoint;
}

// Driver code
int main()
{
    const int N = 6;
    int arr[N][2] = {{-1, 1}, {0, 0}, {1, 1}, {2, 2},
                    {3, 3}, {3, 4}};

    vector< pair<int, int> > points;
    for (int i = 0; i < N; i++)
        points.push_back(make_pair(arr[i][0], arr[i][1]));

    cout << maxPointOnSameLine(points) << endl;

    return 0;
}

Python3

# python3 program to find maximum number of 2D points that lie on the same line.

from collections import defaultdict
from math import gcd
from typing import DefaultDict, List, Tuple

IntPair = Tuple[int, int]


def normalized_slope(a: IntPair, b: IntPair) -> IntPair:
    """
    Returns normalized (rise, run) tuple. We won't return the actual rise/run
    result in order to avoid floating point math, which leads to faulty
    comparisons.

    See
    https://en.wikipedia.org/wiki/Floating-point_arithmetic#Accuracy_problems
    """
    run = b[0] - a[0]

    # normalize undefined slopes to (1, 0)
    if run == 0:
        return (1, 0)

    # normalize to left-to-right
    if run < 0:
        a, b = b, a
        run = b[0] - a[0]

    rise = b[1] - a[1]
    # Normalize by greatest common divisor.
    # math.gcd only works on positive numbers.
    gcd_ = gcd(abs(rise), run)
    return (
        rise // gcd_,
        run // gcd_,
    )


def maximum_points_on_same_line(points: List[List[int]]) -> int:
    # You need at least 3 points to potentially have non-collinear points.
    # For [0, 2] points, all points are on the same line.
    if len(points) < 3:
        return len(points)

    # Note that every line we find will have at least 2 points.
    # There will be at least one line because len(points) >= 3.
    # Therefore, it's safe to initialize to 0.
    max_val = 0

    for a_index in range(0, len(points) - 1):
        # All lines in this iteration go through point a.
        # Note that lines a-b and a-c cannot be parallel.
        # Therefore, if lines a-b and a-c have the same slope, they're the same
        # line.
        a = tuple(points[a_index])
        # Fresh lines already have a, so default=1
        slope_counts: DefaultDict[IntPair, int] = defaultdict(lambda: 1)

        for b_index in range(a_index + 1, len(points)):
            b = tuple(points[b_index])
            slope_counts[normalized_slope(a, b)] += 1

        max_val = max(
            max_val,
            max(slope_counts.values()),
        )

    return max_val


print(maximum_points_on_same_line([
    [-1, 1],
    [0, 0],
    [1, 1],
    [2, 2],
    [3, 3],
    [3, 4],
]))

# This code is contributed by Jose Alvarado Torre

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