Count of Arrays of size N with elements in range [0, (2^K)-1] having maximum sum & bitwise AND 0

Given two integers N and K, The task is to find the count of all possible arrays of size N with maximum sum & bitwise AND of all elements as 0. Also, elements should be within the range of 0 to 2^K-1.

Examples:

Input: N = 3,  K = 2
Output: 9
Explanation:  2² - 1 = 3, so elements of  arrays should be between 0 to 3. All possible arrays are- [3, 3, 0],  [1, 2, 3], [3, 0, 3], [0, 3, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1] Bitwise AND of all the arrays is 0 & also the sum = 6 is maximum 
Input: N = 2, K = 2
Output: 4
Explanation:  All possible arrays are -  [3, 0], [0, 3], [1, 2], [2, 1]

Approach: To better understand the approach, refer to the steps below:

• As the maximum possible element is (2^K - 1) and the size of the array is N so if all elements of the array are equal to the maximum element then the sum will be maximum i.e.  N * (2^K - 1) = N * ( 2⁰ + 2¹ + ................. + 2^{K - 1} ). Keep in mind that there are K bits in ( 2^K - 1) and all bits are set.
• So now to make bitwise AND of all elements equal to 0 we have to unset each bit at least in one element. Also, we can not unset the same bit in more than 1 element because in that case sum will not be maximum.
• After unsetting each bit in one element, the maximum possible Sum = N * ( 2⁰ + 2¹ + ......... + 2^{K - 1} ) - ( 2⁰ + 2¹ + .......... + 2^{K - 1} ) = (N * (2^K -1 )) - (2^K -1)= (N - 1) * (2^K - 1).
• Now the goal is to find all the ways through which we can unset all K bits in at least one element. You can see that for unsetting a single bit you have N options i.e. you can unset it in any one of N elements. So the total way for unsetting K bits will be N^K. This is our final answer.

Illustration:

Let N = 3, K = 3
 

• Make all elements of the array equal to 2³ - 1 = 7. The array will be [7, 7, 7]. Take binary representation of all elements : [111, 111, 111].
• Unset each bit in exactly one element. Suppose we unset the 3rd bit of the 1st element and the 1st two bits of the 2nd element. array becomes [110, 001, 111] = [6, 1, 7]. This is one of the valid arrays. You can generate all arrays in such a way.
• The total number of arrays will be 3³ = 27.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Iterative Function to calculate
// (x^y) in O(log y)
int power(int x, int y)
{

    // Initialize answer
    int res = 1;

    // Check till the number becomes zero
    while (y) {

        // If y is odd, multiply x with result
        if (y % 2 == 1)
            res = (res * x);

        // y = y/2
        y = y >> 1;

        // Change x to x^2
        x = (x * x);
    }
    return res;
}

// Driver Code
int main()
{
    int N = 3, K = 2;
    cout << power(N, K);
    return 0;
}

// Java code for the above approach
import java.util.*;
public class GFG
{

  // Iterative Function to calculate
  // (x^y) in O(log y)
  static int power(int x, int y)
  {

    // Initialize answer
    int res = 1;

    // Check till the number becomes zero
    while (y > 0) {

      // If y is odd, multiply x with result
      if (y % 2 == 1)
        res = (res * x);

      // y = y/2
      y = y >> 1;

      // Change x to x^2
      x = (x * x);
    }
    return res;
  }

  // Driver Code
  public static void main(String args[])
  {
    int N = 3, K = 2;
    System.out.print(power(N, K));
  }
}

// This code is contributed by Samim Hossain Mondal.

# python3 program for the above approach

# Iterative Function to calculate
# (x^y) in O(log y)
def power(x, y):

    # Initialize answer
    res = 1

    # Check till the number becomes zero
    while (y):

        # If y is odd, multiply x with result
        if (y % 2 == 1):
            res = (res * x)

        # y = y/2
        y = y >> 1

        # Change x to x^2
        x = (x * x)

    return res

# Driver Code
if __name__ == "__main__":

    N = 3
    K = 2
    print(power(N, K))

    # This code is contributed by rakeshsahni

// C# code to implement above approach
using System;
class GFG
{
  
  // Iterative Function to calculate
  // (x^y) in O(log y)
  static int power(int x, int y)
  {

    // Initialize answer
    int res = 1;

    // Check till the number becomes zero
    while (y > 0) {

      // If y is odd, multiply x with result
      if (y % 2 == 1)
        res = (res * x);

      // y = y/2
      y = y >> 1;

      // Change x to x^2
      x = (x * x);
    }
    return res;
  }

  // Driver code
  public static void Main()
  {
    int N = 3, K = 2;
    Console.Write(power(N, K));
  }
}

// This code is contributed by Samim Hossain Mondal.

    <script>
        // JavaScript code for the above approach

        // Iterative Function to calculate
        // (x^y) in O(log y)
        function power(x, y) {

            // Initialize answer
            let res = 1;

            // Check till the number becomes zero
            while (y) {

                // If y is odd, multiply x with result
                if (y % 2 == 1)
                    res = (res * x);

                // y = y/2
                y = y >> 1;

                // Change x to x^2
                x = (x * x);
            }
            return res;
        }

        // Driver Code

        let N = 3, K = 2;
        document.write(power(N, K));

  // This code is contributed by Potta Lokesh
    </script>

9

Time Complexity: O(logK)
Auxiliary Space: O(1)