Sum of squares of all Subsets of given Array
Last Updated :
13 Mar, 2022
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++
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long power( int p)
{
long long res = 1;
for ( int i = 1; i <= p; ++i) {
res *= 2;
res %= mod;
}
return res % mod;
}
long long subset_square_sum(vector< int >& A)
{
int n = ( int )A.size();
long long ans = 0;
for ( int i : A) {
ans += (1LL * i * i) % mod;
ans %= mod;
}
return (1LL * ans * power(n - 1)) % mod;
}
int main()
{
vector< int > A = { 3, 7 };
cout << subset_square_sum(A);
return 0;
}
|
Java
class GFG
{
static final int mod = ( int )(1e9 + 7 );
static long power( int p)
{
long res = 1 ;
for ( int i = 1 ; i <= p; ++i)
{
res *= 2 ;
res %= mod;
}
return res % mod;
}
static long subset_square_sum( int A[])
{
int n = A.length;
long ans = 0 ;
for ( int i : A)
{
ans += ( 1 * i * i) % mod;
ans %= mod;
}
return ( 1 * ans * power(n - 1 )) % mod;
}
public static void main (String[] args)
{
int A[] = { 3 , 7 };
System.out.println(subset_square_sum(A));
}
}
|
Python3
mod = 10 * * 9 + 7
def power(p):
res = 1
for i in range ( 1 , p + 1 ):
res * = 2
res % = mod
return res % mod
def subset_square_sum(A):
n = len (A)
ans = 0
for i in A:
ans + = i * i % mod
ans % = mod
return ans * power(n - 1 ) % mod
A = [ 3 , 7 ]
print (subset_square_sum(A))
|
C#
using System;
class GFG
{
static readonly int mod = ( int )(1e9 + 7);
static long power( int p)
{
long res = 1;
for ( int i = 1; i <= p; ++i)
{
res *= 2;
res %= mod;
}
return res % mod;
}
static long subset_square_sum( int []A)
{
int n = A.Length;
long ans = 0;
foreach ( int i in A)
{
ans += (1 * i * i) % mod;
ans %= mod;
}
return (1 * ans * power(n - 1)) % mod;
}
public static void Main (String[] args)
{
int []A = { 3, 7 };
Console.WriteLine(subset_square_sum(A));
}
}
|
Javascript
<script>
const mod = 1000000000 + 7;
function power(p)
{
let res = 1;
for (let i = 1; i <= p; ++i) {
res *= 2;
res %= mod;
}
return res % mod;
}
function subset_square_sum(A)
{
let n = A.length;
let ans = 0;
for (let i = 0; i < n; i++) {
ans += (A[i] * A[i]) % mod;
ans %= mod;
}
return (ans * power(n - 1)) % mod;
}
let A = [ 3, 7 ];
document.write(subset_square_sum(A));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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