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.


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++ program to find maximum number of point
   which lie on same line */
#include <bits/stdc++.h>
using namespace std;

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

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

    // map to store slope pair
    unordered_map<pair<int, 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])

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

                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 colinear point which contains current
        // point are : %d\n", curMax + overlapPoints + 1);

    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;



This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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