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

• Last Updated : 26 Jul, 2022

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 codeint 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 coden = 3print(findF_N(n)) # This code is contributed by# Sanjit_Prasad

## C#

 // C# implementation of above approachusing 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)

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