Maximum distinct lines passing through a single point

Given N    lines represented by two points (x1, y1)    and (x2, y2)    . The task is to find maximum number of lines which can pass through a single point, without superimposing (or covering) any other line. We can move any line but not rotate it.
Examples: 
 

Input : Line 1 : x1 = 1, y1 = 1, x2 = 2, y2 = 2
        Line 2 : x2 = 2, y1 = 2, x2 = 4, y2 = 10
Output : 2
There are two lines. These two lines are not 
parallel, so both of them will pass through
a single point.


Input : Line 1 : x1 = 1, y1 = 5, x2 = 1, y2 = 10
        Line 2 : x2 = 5, y1 = 1, x2 = 10, y2 = 1
Output : 2

• Represent lines as pair (m, c)    where line can be given as y=mx+c    , called line slope form. We can now see that we can change the c for any line, but cannot modify m.
• Lines having same value of m parallel, given that (c1 ? c2). Also no two parallel lines can pass through same point without superimposing to each other.
• So, our problem reduces to finding different values of slopes from given set of lines.

We can calculate slope of a line as \frac{(y2-y1)}{(x2-x1)}    , add them to a set and count the number of distinct values of slope in set. But we have to handle vertical lines separately. 
So, if x1 = x2    then, slope = INT_MAX. 
Otherwise, slope = \frac{(y2-y1)}{(x2-x1)}    .
Below is the implementation of the approach. 
 

// C++ program to find maximum number of lines
// which can pass through a single point
#include <bits/stdc++.h>
using namespace std;

// function to find maximum lines which passes
// through a single point
int maxLines(int n, int x1[], int y1[], 
                int x2[], int y2[])
{
    unordered_set<double> s;

    double slope;
    for (int i = 0; i < n; ++i) {
        if (x1[i] == x2[i])
            slope = INT_MAX;
        else
            slope = (y2[i] - y1[i]) * 1.0 
                    / (x2[i] - x1[i]) * 1.0;

        s.insert(slope);
    }

    return s.size();
}

// Driver program
int main()
{
    int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },
            x2[] = { 2, 4 }, y2[] = { 2, 10 };
    cout << maxLines(n, x1, y1, x2, y2);
    return 0;
}
// This code is written by
// Sanjit_Prasad

// Java program to find maximum number of lines
// which can pass through a single point

import java.util.*;
import java.lang.*;
import java.io.*;

class GFG{

// function to find maximum lines which passes
// through a single point
static int maxLines(int n, int x1[], int y1[], 
                    int x2[], int y2[])
{
    Set<Double> s=new HashSet<Double>();

    double slope;
    for (int i = 0; i < n; ++i) {
        if (x1[i] == x2[i])
            slope = Integer.MAX_VALUE;
        else
            slope = (y2[i] - y1[i]) * 1.0 
                    / (x2[i] - x1[i]) * 1.0;

        s.add(slope);
    }

    return s.size();
}

// Driver program
public static void main(String args[])
{
    int n = 2, x1[] = { 1, 2 }, y1[] = { 1, 2 },
            x2[] = { 2, 4 }, y2[] = { 2, 10 };
    System.out.print(maxLines(n, x1, y1, x2, y2));
}
}
// This code is written by
// Subhadeep

# Python3 program to find maximum number 
# of lines which can pass through a 
# single point
import sys
# function to find maximum lines 
# which passes through a single point
def maxLines(n, x1, y1, x2, y2):

    s = [];

    slope=sys.maxsize;
    for i in range(n):
        if (x1[i] == x2[i]):
            slope = sys.maxsize;
        else:
            slope = (y2[i] - y1[i]) * 1.0 /(x2[i] - x1[i]) * 1.0;

        s.append(slope);

    return len(s);

# Driver Code
n = 2;
x1 = [ 1, 2 ];
y1 = [1, 2];
x2 = [2, 4];
y2 = [2, 10];
print(maxLines(n, x1, y1, x2, y2));

# This code is contributed by mits

// C# program to find maximum number of lines
// which can pass through a single point
using System;
using System.Collections.Generic;

class GFG
{

// function to find maximum lines which passes
// through a single point
static int maxLines(int n, int []x1, int []y1, 
                    int []x2, int []y2)
{
    HashSet<Double> s = new HashSet<Double>();

    double slope;
    for (int i = 0; i < n; ++i) 
    {
        if (x1[i] == x2[i])
            slope = int.MaxValue;
        else
            slope = (y2[i] - y1[i]) * 1.0
                    / (x2[i] - x1[i]) * 1.0;

        s.Add(slope);
    }

    return s.Count;
}

// Driver code
public static void Main()
{
    int n = 2;
    int []x1 = { 1, 2 }; int []y1 = { 1, 2 };
    int []x2 = { 2, 4 }; int []y2 = { 2, 10 };
    Console.Write(maxLines(n, x1, y1, x2, y2));
}
}

/* This code contributed by PrinciRaj1992 */

<?php
// PHP program to find maximum number 
// of lines which can pass through a 
// single point

// function to find maximum lines 
// which passes through a single point
function maxLines($n, $x1, $y1, $x2, $y2)
{
    $s = array();

    $slope;
    for ($i = 0; $i < $n; ++$i)
    {
        if ($x1[$i] == $x2[$i])
            $slope = PHP_INT_MAX;
        else
            $slope = ($y2[$i] - $y1[$i]) * 1.0 /
                     ($x2[$i] - $x1[$i]) * 1.0;

        array_push($s, $slope);
    }

    return count($s);
}

// Driver Code
$n = 2;
$x1 = array( 1, 2 );
$y1 = array(1, 2);
$x2 = array(2, 4);
$y2 = array(2, 10);
echo maxLines($n, $x1, $y1, $x2, $y2);

// This code is contributed by mits
?>

 <script>
      // JavaScript program to find maximum number
      // of lines which can pass through a
      // single point
      // function to find maximum lines
      // which passes through a single point
      function maxLines(n, x1, y1, x2, y2) {
        var s = [];

        //Max Integer Value
        var slope = 2147483647;

        for (let i = 0; i < n; i++) {
          if (x1[i] === x2[i]) slope = 2147483647;
          else slope = (((y2[i] - y1[i]) * 1.0) / (x2[i] - x1[i])) * 1.0;

          s.push(slope);
        }
        return s.length;
      }
      // Driver Code
      var n = 2;
      var x1 = [1, 2];
      var y1 = [1, 2];
      var x2 = [2, 4];
      var y2 = [2, 10];
      document.write(maxLines(n, x1, y1, x2, y2));
    </script>

2

Time Complexity: O(N)   

Space Complexity: O(N) since using auxiliary space for set