Given a set Sn that represents a finite set {1, 2, 3, ….., n}. Zn denotes the set of all subsets of Sn which contains exactly 2 elements. The task is to find the value of F(n).
Examples:
Input: N = 3 Output: 4 For n=3 we get value 1, 2 times and 2, 1 times thus the answer would be 1 * 2 + 2 * 1 = 4. Input: N = 10 Output: 165
For each value as we go from left to right in the set we get each value, n-value number of times that value.
i.e. For each i, value to be added is (i + 1) * (n – i – 1)
Below is the implementation of the above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
#define ll long long // Function to find the value of F(n) ll findF_N(ll n) { ll ans = 0;
for (ll i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
} // Driver code int main()
{ ll n = 3;
cout << findF_N(n);
return 0;
} |
Java
// Java implementation of above approach import java.io.*;
class GFG {
// Function to find the value of F(n)
static long findF_N( long n)
{
long ans = 0 ;
for ( long i = 0 ; i < n; ++i)
ans += (i + 1 ) * (n - i - 1 );
return ans;
}
// Driver code
public static void main(String[] args)
{
long n = 3 ;
System.out.println(findF_N(n));
}
} // This code is contributed by anuj_67.. |
Python3
# Python3 implementation of # above approach # Function to find the value of F(n) def findF_N(n):
ans = 0
for i in range (n):
ans = ans + (i + 1 ) * (n - i - 1 )
return ans
# Driver code n = 3
print (findF_N(n))
# This code is contributed by # Sanjit_Prasad |
C#
// C# implementation of above approach using System;
class GFG {
// Function to find the
// value of F(n)
static long findF_N( long n)
{
long ans = 0;
for ( long i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
}
// Driver code
public static void Main()
{
long n = 3;
Console.WriteLine(findF_N(n));
}
} // This code is contributed by anuj_67 |
PHP
<?php // PHP implementation of above approach // Function to find the value of F(n) function findF_N( $n )
{ $ans = 0;
for ( $i = 0; $i < $n ; ++ $i )
$ans += ( $i + 1) * ( $n - $i - 1);
return $ans ;
} // Driver code $n = 3;
echo findF_N( $n );
// This code is contributed // by Akanksha Rai(Abby_akku) |
Javascript
<script> // JavaScript implementation of above approach // Function to find the value of F(n) function findF_N(n)
{ var ans = 0;
for ( var i = 0; i < n; ++i)
ans += (i + 1) * (n - i - 1);
return ans;
} // Driver code var n = 3;
document.write( findF_N(n)); </script> |
Output:
4
Time Complexity: O(n) // n is the length of the array
Space Complexity: O(1)