Find the sum of power of bit count raised to the power B

Given an integer, array A. Find the sum of set bits raised to the power A[i] for each element in A[i].

Example:

Input: N = 3, A[] = {1, 2, 3}
Output: 10
Explanation:
Set bit of each array element is
1 = 1 set bit, 
2 = 1 set bit, 
3 = 2 set bit 
store each set bit in b[i].
Compute sum of power(b[i], i) 
where i is ranging from 1 to n.
that is sum = power(1, 1)+
power(1, 2)+power(2, 3) = 10

Input: N = 4, A[] = {2, 4, 5, 3}
Output: 42

Approach:
We can use modular-exponentiation method to calculate a^b under mod m and inbuilt __builtin_popcount function to count number of set bits in binary representation of A[i].



Below is the implementation of the above approach.

C++

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// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate a^b mod m
// using fast-exponentiation method
int fastmod(int base, int exp, int mod)
{
    if (exp == 0)
      return 1;
    else if (exp % 2 == 0) {
      int ans = fastmod(base, exp / 2, mod);
      return (ans % mod * ans % mod) % mod;
    }
    else
      return (fastmod(base, exp - 1, mod) % mod * base % mod) % mod;
}
  
// Function to
// calculate sum
int findPowerSum(int n, int ar[])
{
  
    const int mod = 1e9 + 7;
    int sum = 0;
  
    // Itereate for all
    // values of array A
    for (int i = 0; i < n; i++) {
        int base = __builtin_popcount(ar[i]);
        int exp = ar[i];
        // Calling fast-exponentiation and
        // appending ans to sum
        sum += fastmod(base, exp, mod);
        sum %= mod;
    }
  
    return sum;
}
  
// Driver code.
int main()
{
    int n = 3;
    int ar[] = { 1, 2, 3 };
    cout << findPowerSum(n, ar);
    return 0;
}

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Java

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// Java program for the above approach 
class GFG 
{
      
    // Function to calculate a^b mod m 
    // using fast-exponentiation method 
    static int fastmod(int base, int exp, int mod) 
    
        if (exp == 0
        return 1
        else if (exp % 2 == 0)
        
            int ans = fastmod(base, exp / 2, mod); 
            return (ans % mod * ans % mod) % mod; 
        
        else
            return (fastmod(base, exp - 1, mod) %
                        mod * base % mod) % mod; 
    
      
    /* Function to get no of set 
    bits in binary representation 
    of positive integer n */
    static int countSetBits(int n) 
    
        int count = 0
        while (n > 0)
        
            count += n & 1
            n >>= 1
        
        return count; 
    
      
    // Function to calculate sum 
    static int findPowerSum(int n, int ar[]) 
    
      
        final int mod = (int)1e9 + 7
        int sum = 0
      
        // Itereate for all 
        // values of array A 
        for (int i = 0; i < n; i++) 
        
            int base = countSetBits(ar[i]); 
            int exp = ar[i]; 
              
            // Calling fast-exponentiation and 
            // appending ans to sum 
            sum += fastmod(base, exp, mod); 
            sum %= mod; 
        
      
        return sum; 
    
      
    // Driver code. 
    public static void main (String[] args)
    
        int n = 3
        int ar[] = { 1, 2, 3 }; 
        System.out.println(findPowerSum(n, ar)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 program for the above approach 
  
# Function to calculate a^b mod m 
# using fast-exponentiation method 
def fastmod(base, exp, mod) :
  
    if (exp == 0) : 
        return 1
          
    elif (exp % 2 == 0) :
        ans = fastmod(base, exp / 2, mod); 
          
        return (ans % mod * ans % mod) % mod; 
      
    else :
        return (fastmod(base, exp - 1, mod)
                % mod * base % mod) % mod; 
  
# Function to 
# calculate sum 
def findPowerSum(n, ar) :
      
    mod = int(1e9) + 7;
    sum = 0;
      
    # Itereate for all values of array A
    for i in range(n) :
        base = bin(ar[i]).count('1');
        exp = ar[i];
          
        # Calling fast-exponentiation and 
        # appending ans to sum
        sum += fastmod(base, exp, mod);
        sum %= mod;
          
    return sum
  
# Driver code. 
if __name__ == "__main__"
  
    n = 3
    ar = [ 1, 2, 3 ]; 
      
    print(findPowerSum(n, ar)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# program for the above approach 
using System;
  
class GFG 
{
      
    // Function to calculate a^b mod m 
    // using fast-exponentiation method 
    static int fastmod(int baseval, int exp, int mod) 
    
        if (exp == 0) 
            return 1; 
        else if (exp % 2 == 0)
        
            int ans = fastmod(baseval, exp / 2, mod); 
            return (ans % mod * ans % mod) % mod; 
        
        else
            return (fastmod(baseval, exp - 1, mod) %
                        mod * baseval % mod) % mod; 
    
      
    /* Function to get no of set 
    bits in binary representation 
    of positive integer n */
    static int countSetBits(int n) 
    
        int count = 0; 
        while (n > 0)
        
            count += n & 1; 
            n >>= 1; 
        
        return count; 
    
      
    // Function to calculate sum 
    static int findPowerSum(int n, int []ar) 
    
      
        int mod = (int)1e9 + 7; 
        int sum = 0; 
      
        // Itereate for all 
        // values of array A 
        for (int i = 0; i < n; i++) 
        
            int baseval = countSetBits(ar[i]); 
            int exp = ar[i]; 
              
            // Calling fast-exponentiation and 
            // appending ans to sum 
            sum += fastmod(baseval, exp, mod); 
            sum %= mod; 
        
      
        return sum; 
    
      
    // Driver code. 
    public static void Main ()
    
        int n = 3; 
        int []ar = { 1, 2, 3 }; 
        Console.WriteLine(findPowerSum(n, ar)); 
    
}
  
// This code is contributed by AnkitRai01

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Output:

10

Time Complexity: O(n*log(n))



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Improved By : AnkitRai01