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Count of N-size maximum sum Arrays with elements in range [0, 2^K – 1] and Bitwise AND equal to 0

  • Last Updated : 16 Jul, 2021

Given two positive integers N and K, the task is to find the number of arrays of size N such that each array element lies over the range [0, 2K – 1] with the maximum sum of array element having Bitwise AND of all array elements 0.

Examples:

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Input: N = 2 K = 2
Output: 4
Explanation:
The possible arrays with maximum sum having the Bitwise AND of all array element as 0 {0, 3}, {3, 0}, {1, 2}, {2, 1}. The count of such array is 4.



Input: N = 5 K = 6 
Output: 15625 

Approach: The given problem can be solved by observing the fact that as the Bitwise AND of the generated array should be 0, then for each i in the range [0, K – 1] there should be at least 1 element with an ith bit equal to 0 in its binary representation. Therefore, to maximize the sum of the array, it is optimal to have exactly 1 element with the ith bit unset.

Hence, for each of the K bits, there are NC1 ways to make it unset in 1 array element. Therefore, the resultant count of an array having the maximum sum is given by NK.

Below is the implementation of the approach :

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the value of X to
// the power Y
int power(int x, unsigned int y)
{
    // Stores the value of X^Y
    int res = 1;
 
    while (y > 0) {
 
        // If y is odd, multiply x
        // with result
        if (y & 1)
            res = res * x;
 
        // Update the value of y and x
        y = y >> 1;
        x = x * x;
    }
 
    // Return the result
    return res;
}
 
// Function to count number of arrays
// having element over the range
// [0, 2^K - 1] with Bitwise AND value
// 0 having maximum possible sum
void countArrays(int N, int K)
{
    // Print the value of N^K
    cout << int(power(N, K));
}
 
// Driver Code
int main()
{
    int N = 5, K = 6;
    countArrays(N, K);
 
    return 0;
}

Java




// Java program for the above approach
public class GFG
{
 
// Function to find the value of X to
// the power Y
static int power(int x, int y)
{
   
    // Stores the value of X^Y
    int res = 1;
 
    while (y > 0) {
 
        // If y is odd, multiply x
        // with result
        if (y%2!=0)
            res = res * x;
 
        // Update the value of y and x
        y = y >> 1;
        x = x * x;
    }
 
    // Return the result
    return res;
}
 
// Function to count number of arrays
// having element over the range
// [0, 2^K - 1] with Bitwise AND value
// 0 having maximum possible sum
static void countArrays(int N, int K)
{
   
    // Print the value of N^K
   System.out.println((int)(power(N, K)));
}
 
 
// Driver Code
public static void main(String args[])
{
    int N = 5, K = 6;
    countArrays(N, K);
}
}
 
// This code is contributed by SoumikMondal

Python3




# Python Program for the above approach
 
# Function to find the value of X to
# the power Y
def power(x, y):
    # Stores the value of X^Y
    res = 1
 
    while (y > 0):
 
        # If y is odd, multiply x
        # with result
        if (y & 1):
            res = res * x
 
        # Update the value of y and x
        y = y >> 1
        x = x * x
 
    # Return the result
    return res
 
# Function to count number of arrays
# having element over the range
# [0, 2^K - 1] with Bitwise AND value
# 0 having maximum possible sum
def countArrays(N, K):
    # Print the value of N^K
    print(power(N, K))
 
 
# Driver Code
 
N = 5;
K = 6;
countArrays(N, K)
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the value of X to
// the power Y
static int power(int x, int y)
{
     
    // Stores the value of X^Y
    int res = 1;
 
    while (y > 0)
    {
         
        // If y is odd, multiply x
        // with result
        if (y % 2 != 0)
            res = res * x;
 
        // Update the value of y and x
        y = y >> 1;
        x = x * x;
    }
 
    // Return the result
    return res;
}
 
// Function to count number of arrays
// having element over the range
// [0, 2^K - 1] with Bitwise AND value
// 0 having maximum possible sum
static void countArrays(int N, int K)
{
     
    // Print the value of N^K
    Console.WriteLine((int)(power(N, K)));
}
 
// Driver Code
public static void Main()
{
    int N = 5, K = 6;
     
    countArrays(N, K);
}
}
 
// This code is contributed by subhammahato348

Javascript




<script>
        // JavaScript Program for the above approach
 
 
        // Function to find the value of X to
        // the power Y
        function power(x, y) {
            // Stores the value of X^Y
            let res = 1;
 
            while (y > 0) {
 
                // If y is odd, multiply x
                // with result
                if (y & 1)
                    res = res * x;
 
                // Update the value of y and x
                y = y >> 1;
                x = x * x;
            }
 
            // Return the result
            return res;
        }
 
        // Function to count number of arrays
        // having element over the range
        // [0, 2^K - 1] with Bitwise AND value
        // 0 having maximum possible sum
        function countArrays(N, K) {
            // Print the value of N^K
            document.write(power(N, K));
        }
 
        // Driver Code
 
        let N = 5, K = 6;
        countArrays(N, K);
 
    // This code is contributed by Potta Lokesh
 
    </script>
Output: 
15625

 

Time Complexity: O(log K)
Auxiliary Space: O(1)




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