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

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

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, 2^K – 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 i^th 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 i^th 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 N^K.

Below is the implementation of the approach :

C++

#include <bits/stdc++.h> using namespace std;    int power(int x, unsigned int y) {          int res = 1;        while (y > 0) {                              if (y & 1)             res = res * x;                     y = y >> 1;         x = x * x;     }             return res; }    void countArrays(int N, int K) {          cout << int(power(N, K)); }    int main() {     int N = 5, K = 6;     countArrays(N, K);        return 0; }

Output:

15625

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

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