Count maximum points on same line

4.2

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++ 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])
                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 colinear 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;
}

Output:

4

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

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