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

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  • Last Updated : 16 Jun, 2022
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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 2K-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 (2K – 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 * (2K – 1) = N * ( 20 + 21 + …………….. + 2K – 1 ). Keep in mind that there are K bits in ( 2K – 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 + ……… + 2K – 1 ) – ( 20 + 21 + ………. + 2K – 1 ) = (N * (2K -1 )) – (2K -1)= (N – 1) * (2K – 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 NK. 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++




// 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




// 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




# 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#




// 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.

Javascript




  <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>

Output

9

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


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