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 subsetslong 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 codeint 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 approachmod = 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 subsetsdef 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 codeA = [3, 7]print(subset_square_sum(A))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// Javascript implementation of the approachconst mod = 1000000000 + 7;// Function to return (2^P % mod)function power(p){ let res = 1; for (let i = 1; i <= p; ++i) { res *= 2; res %= mod; } return res % mod;}// Function to return the sum of squares of subsetsfunction subset_square_sum(A){ let n = A.length; let ans = 0; // Sqauaring the elements // and adding it to ans for (let i = 0; i < n; i++) { ans += (A[i] * A[i]) % mod; ans %= mod; } return (ans * power(n - 1)) % mod;}// Driver code let A = [ 3, 7 ]; document.write(subset_square_sum(A));</script> |
116
Time Complexity: O(N)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
