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:
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 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
# 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# 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
<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>
Time Complexity: O(log K)
Auxiliary Space: O(1)
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