Minimum lines to cover all points - GeeksforGeeks

Minimum lines to cover all points - GeeksforGeeks
Given N points in 2-dimensional space, we need to print the count of the minimum number of lines which traverse through all these N points and which go through a specific (xO, yO) point also.

We can solve this problem by considering the slope of all points with (xO, yO). If two distinct points have the same slope with (xO, yO) then they can be covered with same line only so we can track slope of each point and whenever we get a new slope we will increase our line count by one.
In below code slope is stored as a pair of integer to get rid of the precision problem and a set is used to keep track of occurred slopes.

int gcd(int a, int b)

{

    if (b == 0)

        return a;

    return gcd(b, a % b);

}

 

//  method returns reduced form of dy/dx as a pair

pair<int, int> getReducedForm(int dy, int dx)

{

    int g = gcd(abs(dy), abs(dx));

 

    //  get sign of result

    bool sign = (dy < 0) ^ (dx < 0);

 

    if (sign)

        return make_pair(-abs(dy) / g, abs(dx) / g);

    else

        return make_pair(abs(dy) / g, abs(dx) / g);

}

 

/*  method returns minimum number of lines to

    cover all points where all lines goes

    through (xO, yO) */

int minLinesToCoverPoints(int points[][2], int N,

                                   int xO, int yO)

{

    //  set to store slope as a pair

    set< pair<int, int> > st;

    pair<int, int> temp;

    int minLines = 0;

 

    //  loop over all points once

    for (int i = 0; i < N; i++)

    {

        //  get x and y co-ordinate of current point

        int curX = points[i][0];

        int curY = points[i][1];

 

        temp = getReducedForm(curY - yO, curX - xO);

 

        // if this slope is not there in set,

        // increase ans by 1 and insert in set

        if (st.find(temp) == st.end())

        {

            st.insert(temp);

            minLines++;

        }

    }

 

    return minLines;

}

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