# Minimum value possible of a given function from the given set

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  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

 

## Javascript

 

Output:

4`

Time Complexity: O(n) // n is the length of the array

Space Complexity: 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 geeks!

Previous
Next