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Minimum value possible of a given function from the given set

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  • Last Updated : 26 Jul, 2022
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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).

F(n) = \sum_{X\in Z_n }min(X)

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)


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