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++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to count maximum number of // intersections possibleint countIntersections(int n){ return n * (n - 1) / 2;}// Driver Codeint main(){ int n = 3; cout << countIntersections(n); return 0;} |
Java
// 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
# Python3 implementation of the above approach# Function to count maximum number of# intersections possibledef countIntersections(n): return n * (n - 1) // 2# Driver Coden = 3print(countIntersections(n))# This code is contributed by mohit kumar |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the above approach// Function to count maximum number of // intersections possiblefunction countIntersections(n){ return n * (n - 1) / 2;}// Driver Codevar n = 3;document.write(countIntersections(n));</script> |
3
Time Complexity: O(1)
Auxiliary Space: O(1)
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