Given N planes. The task is to find the maximum number of line intersections that can be formed through the intersections of N planes.
Examples:
Input: N = 3
Output: 3
Input: N = 5
Output: 10
Approach:
Let there be N planes such that no 3 planes intersect in a single line of intersection and no 2 planes are parallel to each other. Adding ‘N+1’th plane to this space should be possible while retaining the above two conditions. In that case, this plane would intersect each of the N planes in ‘N’ distinct lines.
Thus, the ‘N+1’th plane could atmost add ‘N’ new lines to the total count of lines of intersection. Similarly, the Nth plane could add at most “N-1′ distinct lines of intersection. It is easy therefore to see that, for ‘N’ planes, the maximum number of lines of intersection could be:
(N-1) + (N-2) +...+ 1 = N*(N-1)/2
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to count maximum number of // intersections possible int countIntersections( int n)
{ return n * (n - 1) / 2;
} // Driver Code int main()
{ int n = 3;
cout << countIntersections(n);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to count maximum number of
// intersections possible
static int countIntersections( int n)
{
return n * (n - 1 ) / 2 ;
}
// Driver Code
public static void main (String[] args)
{
int n = 3 ;
System.out.println(countIntersections(n));
}
} // This code is contributed by AnkitRai01 |
# Python3 implementation of the above approach # Function to count maximum number of # intersections possible def countIntersections(n):
return n * (n - 1 ) / / 2
# Driver Code n = 3
print (countIntersections(n))
# This code is contributed by mohit kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to count maximum number of
// intersections possible
static int countIntersections( int n)
{
return n * (n - 1) / 2;
}
// Driver Code
public static void Main (String[] args)
{
int n = 3;
Console.WriteLine(countIntersections(n));
}
} // This code is contributed by 29AjayKumar |
<script> // Javascript implementation of the above approach // Function to count maximum number of // intersections possible function countIntersections(n)
{ return n * (n - 1) / 2;
} // Driver Code var n = 3;
document.write(countIntersections(n)); </script> |
3
Time Complexity: O(1)
Auxiliary Space: O(1)