Maximum points of intersection n lines

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You are given n straight lines. You have to find a maximum number of points of intersection with these n lines.
Examples:

Input : n = 4
Output : 6

Input : n = 2
Output :1

Approach :
As we have n number of line, and we have to find the maximum point of intersection using this n line. So this can be done using the combination. This problem can be thought of as a number of ways to select any two lines among n line. As every line intersects with others that are selected.
So, the total number of points = nC2

Below is the implementation of the above approach:

C++

// CPP program to find maximum intersecting
// points
#include <bits/stdc++.h>
using namespace std;
#define ll long int
 
 
// nC2 = (n)*(n-1)/2;
ll countMaxIntersect(ll n)
{
   return (n) * (n - 1) / 2;
}
 
// Driver code
int main()
{
    // n is number of line
    ll n = 8;
    cout << countMaxIntersect(n) << endl;
    return 0;
}

Java

// Java program to find maximum intersecting
// points
 
public class GFG {
     
    // nC2 = (n)*(n-1)/2;
    static long countMaxIntersect(long n)
    {
       return (n) * (n - 1) / 2;
    }
 
     
    // Driver code
    public static void main(String args[])
    {
        // n is number of line
        long n = 8;
        System.out.println(countMaxIntersect(n));
 
 
    }
    // This code is contributed by ANKITRAI1
}

Python3

# Python3 program to find maximum 
# intersecting points
 
#nC2 = (n)*(n-1)/2
def countMaxIntersect(n):
    return int(n*(n - 1)/2)
 
#Driver code
if __name__=='__main__':
     
# n is number of line
    n = 8
    print(countMaxIntersect(n))
 
# this code is contributed by 
# Shashank_Sharma

C#

// C# program to find maximum intersecting
// points
using System;
 
class GFG 
{
     
    // nC2 = (n)*(n-1)/2;
    public static long countMaxIntersect(long n)
    {
    return (n) * (n - 1) / 2;
    }
 
     
    // Driver code
    public static void Main()
    {
        // n is number of line
        long n = 8;
        Console.WriteLine(countMaxIntersect(n));
    }
}
// This code is contributed by Soumik

PHP

<?PHP
// PHP program to find maximum intersecting 
// points
 
// nC2 = (n)*(n-1)/2; 
function countMaxIntersect($n) 
{ 
    return ($n) * ($n - 1) / 2; 
} 
 
// Driver code 
 
// n is number of line 
$n = 8;
echo countMaxIntersect($n) . "\n"; 
 
// This code is contributed by ChitraNayal
?>

Javascript

<script>
 
// Javascript program to find maximum intersecting
// points
 
// nC2 = (n)*(n-1)/2;
function countMaxIntersect(n)
{
   return (n) * (n - 1) / 2;
}
 
// Driver code
 
// n is number of line
var n = 8;
document.write( countMaxIntersect(n) );
 
</script>

Output:

Time Complexity: O(1)
Auxiliary Space: O(1)

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