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Count maximum points on same line

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  • Difficulty Level : Hard
  • Last Updated : 20 Apr, 2022

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
    """
    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


This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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