Maximum distinct lines passing through a single point

Last Updated : 28 Jul, 2022

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++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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>

Output:

Time Complexity: $O(N)$

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

Find the maximum possible distance from origin using given points

Sanjit_Prasad

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Maximum distinct lines passing through a single point

C++

Java

Python3

C#

PHP

Javascript

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