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 Yint 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 sumvoid countArrays(int N, int K){ // Print the value of N^K cout << int(power(N, K));}// Driver Codeint main(){ int N = 5, K = 6; countArrays(N, K); return 0;} |
Java
// Java program for the above approachpublic class GFG{// Function to find the value of X to// the power Ystatic 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 sumstatic void countArrays(int N, int K){ // Print the value of N^K System.out.println((int)(power(N, K)));}// Driver Codepublic 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 Ydef 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 sumdef countArrays(N, K): # Print the value of N^K print(power(N, K))# Driver CodeN = 5;K = 6;countArrays(N, K)# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;class GFG{// Function to find the value of X to// the power Ystatic 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 sumstatic void countArrays(int N, int K){ // Print the value of N^K Console.WriteLine((int)(power(N, K)));}// Driver Codepublic 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> |
15625
Time Complexity: O(log K)
Auxiliary Space: O(1)
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