Given an array
arr[]
, the task is to calculate the sum of cubes of all possible non-empty subsets of the given array. Since, the answer can be large, print the value as
mod
1000000007.
Examples:
Input: arr[] = {1, 2} Output: 18 subset({1}) = 13 = 1 subsetval({2}) = 23 = 8 subset({1, 2}) = 13 + 23 = 1 + 8 = 9 Sum of cubes of all Subsets = 1 + 8 + 9 = 18 Input: arr[] = {1, 1, 1} Output: 12
Naive approach:
A simple approach is to
and then cube each element in that subset and add it to the result. The time complexity of this approach will be
O(2
N
)
Efficient approach:
- It can be observed that each element of the original array appears in 2(N – 1) times in all subsets.
- Therefore contribution of any element arri in the final answer will be
arri * 2(N – 1)
- So, the Sum of cubes of all Subsets will be
[arr03 + arr13 + arr23 + … + arr(N-1)3] * 2(N – 1)
Below is the implementation of the above approach:
C++
// C++ implementation of the approachÂ
#include <bits/stdc++.h>using namespace std;Â
const int mod = 1e9 + 7;Â
// Function to return (2^P % mod)long long power(int p){Â Â Â Â long long res = 1;Â Â Â Â for (int i = 1; i <= p; ++i) {Â Â Â Â Â Â Â Â res *= 2;Â Â Â Â Â Â Â Â res %= mod;Â Â Â Â }Â Â Â Â return res % mod;}Â
// Function to return// the sum of cubes of subsetslong long subset_cube_sum(vector<int>& A){Â
    int n = (int)A.size();Â
    long long ans = 0;Â
    // cubing the elements    // and adding it to ans    for (int i : A) {        ans += (1LL * i * i * i) % mod;        ans %= mod;    }Â
    return (1LL * ans * power(n - 1))           % mod;}Â
// Driver codeint main(){Â Â Â Â vector<int> A = { 1, 2 };Â
    cout << subset_cube_sum(A);Â
    return 0;} |
Java
// Java implementation of the approachpublic class GFG {Â Â Â Â static final long MOD = (long) 1e9 + 7;Â
    // 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;    }Â
    // Function to return the sum of cubes of subsets    static long subsetCubeSum(int[] A) {        int n = A.length;        long ans = 0;Â
        // Cubing the elements and adding it to ans        for (int i : A) {            ans += (i * i * i) % MOD;            ans %= MOD;        }Â
        return (ans * power(n - 1)) % MOD;    }Â
    // Driver code    public static void main(String[] args) {        int[] A = {1, 2};        System.out.println(subsetCubeSum(A));    }} |
Python3
# Python3 implementation of the approach mod = int(1e9) + 7; Â
# Function to return (2^P % mod) def power(p) :Â
    res = 1;     for i in range(1, p + 1) :        res *= 2;         res %= mod;          return res % mod; Â
# Function to return # the sum of cubes of subsets def subset_cube_sum(A) : Â
    n = len(A); Â
    ans = 0; Â
    # cubing the elements     # and adding it to ans     for i in A :        ans += (i * i * i) % mod;         ans %= mod; Â
    return (ans * power(n - 1)) % mod; Â
# Driver code if __name__ == "__main__" : Â
    A = [ 1, 2 ]; Â
    print(subset_cube_sum(A));      # This code is contributed by Yash_R |
C#
// C# implementation for the above approachusing System;Â
public class GFG {Â Â Â Â static readonly long MOD = (long)1e9 + 7;Â
    // Function to return (2^P % MOD)    static long Power(int p) {        long res = 1;        for (int i = 1; i <= p; i++) {            res = (res * 2) % MOD;        }        return res;    }Â
    // Function to return the sum of cubes of subsets    static long SubsetCubeSum(int[] A) {        int n = A.Length;        long ans = 0;Â
        // Cubing the elements and adding it to ans        foreach (int i in A) {            ans = (ans + ((i * i * i) % MOD)) % MOD;        }Â
        return (ans * Power(n - 1)) % MOD;    }Â
    // Driver code    public static void Main(string[] args) {        int[] A = { 1, 2 };        Console.WriteLine(SubsetCubeSum(A));    }} |
Output
18
Time Complexity:
O(N)
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