# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program for the ` `// above approach ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```10
```

Time Complexity: O(n*log(n)) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : AnkitRai01