Maximum points of intersections possible among X circles and Y straight lines

0 0 3 minutes read

Given two integers X and Y, the task is to find the maximum number of points of intersection possible among X circles and Y straight lines.

Example:

Input: X = 4, Y = 4
Output: 50
Explanation:
4 lines intersect each other at 6 points and 4 circles intersect each other at maximum of 12 points.
Each line intersects 4 circles at 8 points.
Hence, 4 lines intersect four circles at a maximum of 32 points.
Thus, required number of intersections = 6 + 12 + 32 = 50.

Input: X = 3, Y = 4
Output: 36

Approach:
It can be observed that there are three types of intersections:

The number of ways to choose a pair of points from X circles is $\binom{X}{2}$ . Each such pair intersect at most two points.
The number of ways to choose a pair of points from Y lines is $\binom{Y}{2}$ . Each such pair intersect in at most one point.
The number of ways to choose one circle and one line from X circles and Y lines is $X*Y$ . Each such pair intersect in at most two points.

So, the maximum number of point of intersection can be calculated as:
=> $2*\binom{X}{2} + \binom{Y}{2} + 2*X*Y$
=> $X*(X - 1) + \frac{Y*(Y - 1)}{2} + 2*X*Y$

Thus, formula to find maximum number of point of intersection of X circles and Y straight lines is:
$\frac{Y*(Y - 1)}{2} + X*(2*Y + X - 1)$

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
int maxPointOfIntersection(int x, int y) 
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
int main()
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    cout << (maxPointOfIntersection(x, y));
}
 
// This code is contributed by Ritik Bansal

Java

// Java program to implement
// the above approach
import java.io.*;
public class GFG{
     
static int maxPointOfIntersection(int x, int y) 
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    System.out.print(maxPointOfIntersection(x, y));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to implement
# the above approach
def maxPointOfIntersection(x, y):
    k = y * ( y - 1 ) // 2
    k = k + x * ( 2 * y + x - 1 )
    return k 
 
# Number of circles
x = 3
# Number of straight lines
y = 4
 
# Function Call
print(maxPointOfIntersection(x, y))

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
static int maxPointOfIntersection(int x, int y) 
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    Console.Write(maxPointOfIntersection(x, y));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
// Javascript program to implement 
// the above approach 
function maxPointOfIntersection(x, y) 
{
    let k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
 
// Number of circles
let x = 3;
   
// Number of straight lines
let y = 4;
 
// Function Call
document.write(maxPointOfIntersection(x, y));
 
// This code is contributed by rameshtravel07
 
</script>

Output:

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!