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Sum of squares of all Subsets of given Array
  • Last Updated : 09 Oct, 2019

Given an array arr[]. The value of a subset of array A is defined as the sum of squares of all the numbers in that subset. The task is to calculate the sum of values of all possible non-empty subsets of the given array.
Since, the answer can be large print the val mod 1000000007.

Examples:

Input: arr[] = {3, 7}
Output: 116
val({3}) = 32 = 9
val({7}) = 72 = 49
val({3, 7}) = 32 + 72 = 9 + 49 = 58
9 + 49 + 58 = 116

Input: arr[] = {1, 1, 1}
Output: 12

Naive approach: A simple approach is to find all the subset and then square each element in that subset and add it to the result. The time complexity of this approach will be O(2N)



Efficient approach: It can be observed that in all the possible subsets of the given array, every element will occur 2N – 1 times where N is the size of the array.
So the contribution of any element X in the sum will be 2N – 1 * X2.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int mod = 1e9 + 7;
  
// Function to return (2^P % mod)
long long power(int p)
{
    long long res = 1;
    for (int i = 1; i <= p; ++i) {
        res *= 2;
        res %= mod;
    }
    return res % mod;
}
  
// Function to return the sum of squares of subsets
long long subset_square_sum(vector<int>& A)
{
  
    int n = (int)A.size();
  
    long long ans = 0;
  
    // Sqauaring the elements
    // and adding it to ans
    for (int i : A) {
        ans += (1LL * i * i) % mod;
        ans %= mod;
    }
  
    return (1LL * ans * power(n - 1)) % mod;
}
  
// Driver code
int main()
{
    vector<int> A = { 3, 7 };
  
    cout << subset_square_sum(A);
  
    return 0;
}

Java




// Java implementation of the approach 
class GFG 
{
    static final int mod = (int)(1e9 + 7); 
      
    // Function to return (2^P % mod) 
    static long power(int p) 
    
        long res = 1
        for (int i = 1; i <= p; ++i)
        
            res *= 2
            res %= mod; 
        
        return res % mod; 
    
      
    // Function to return the sum of squares of subsets 
    static long subset_square_sum(int A[]) 
    
        int n = A.length; 
      
        long ans = 0
      
        // Sqauaring the elements 
        // and adding it to ans 
        for (int i : A) 
        
            ans += (1 * i * i) % mod; 
            ans %= mod; 
        
        return (1 * ans * power(n - 1)) % mod; 
    
      
    // Driver code 
    public static void main (String[] args)
    
        int A[] = { 3, 7 }; 
      
        System.out.println(subset_square_sum(A)); 
    
}
  
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
mod = 10**9 + 7
  
# Function to return (2^P % mod)
def power(p):
  
    res = 1
    for i in range(1, p + 1):
        res *= 2
        res %= mod
  
    return res % mod
  
# Function to return the sum of
# squares of subsets
def subset_square_sum(A):
  
    n = len(A)
  
    ans = 0
  
    # Squaring the elements
    # and adding it to ans
    for i in A:
        ans += i * i % mod
        ans %= mod
  
    return ans * power(n - 1) % mod
  
# Driver code
A = [3, 7]
  
print(subset_square_sum(A))
  
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
      
class GFG 
{
    static readonly int mod = (int)(1e9 + 7); 
      
    // Function to return (2^P % mod) 
    static long power(int p) 
    
        long res = 1; 
        for (int i = 1; i <= p; ++i)
        
            res *= 2; 
            res %= mod; 
        
        return res % mod; 
    
      
    // Function to return the sum of squares of subsets 
    static long subset_square_sum(int []A) 
    
        int n = A.Length; 
      
        long ans = 0; 
      
        // Sqauaring the elements 
        // and adding it to ans 
        foreach (int i in A) 
        
            ans += (1 * i * i) % mod; 
            ans %= mod; 
        
        return (1 * ans * power(n - 1)) % mod; 
    
      
    // Driver code 
    public static void Main (String[] args)
    
        int []A = { 3, 7 }; 
      
        Console.WriteLine(subset_square_sum(A)); 
    
}
      
// This code is contributed by 29AjayKumar
Output:
116

Time Complexity: O(N)

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