按位 AND 是 2 的幂的对的计数

给定一个由N 个正整数组成的数组arr[] 。任务是找到按位与值是 2 的幂的对的数量。
例子：

Input: arr[] = {2, 1, 3, 4} 
Output: 2 
Explanation: 
There are 2 pairs (2, 3) and (1, 3) in this array whose Bitwise AND values are: 
1. (2 & 3) = 1 = (2⁰) 
2. (1 & 3) = 1 = (2⁰).
Input: arr[] = {6, 4, 2, 3} 
Output: 4 
Explanation: 
There are 4 pairs (6, 4), (6, 2), (6, 3), (2, 3) whose Bitwise and is power of 2. 
 

方法：对于给定数组中的每个可能对，检查每对元素的按位与是否是2 的完美幂的想法。如果“是”，则计算这一对否则检查下一对。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if x is power of 2
bool check(int x)
{
    // Returns true if x is a power of 2
    return x && (!(x & (x - 1)));
}
 
// Function to return the
// number of valid pairs
int count(int arr[], int n)
{
    int cnt = 0;
 
    // Iterate for all possible pairs
    for (int i = 0; i < n - 1; i++) {
 
        for (int j = i + 1; j < n; j++) {
 
            // Bitwise and value of
            // the pair is passed
            if (check(arr[i]
                      & arr[j]))
                cnt++;
        }
    }
 
    // Return the final count
    return cnt;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 6, 4, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << count(arr, n);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Method to check if x is power of 2
static boolean check(int x)
{
 
    // First x in the below expression
    // is for the case when x is 0
    return x != 0 && ((x & (x - 1)) == 0);
}
 
// Function to return the
// number of valid pairs
static int count(int arr[], int n)
{
    int cnt = 0;
 
    // Iterate for all possible pairs
    for(int i = 0; i < n - 1; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
           
          // Bitwise and value of
          // the pair is passed
          if (check(arr[i] & arr[j]))
              cnt++;
       }
    }
     
    // Return the final count
    return cnt;
}
 
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = new int[]{ 6, 4, 2, 3 };
 
    int n = arr.length;
     
    // Function call
    System.out.print(count(arr, n));
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
 
# Function to check if x is power of 2
def check(x):
     
    # Returns true if x is a power of 2
    return x and (not(x & (x - 1)))
 
# Function to return the
# number of valid pairs
def count(arr, n):
     
    cnt = 0
 
    # Iterate for all possible pairs
    for i in range(n - 1):
        for j in range(i + 1, n):
 
            # Bitwise and value of
            # the pair is passed
            if check(arr[i] & arr[j]):
                cnt = cnt + 1
 
    # Return the final count
    return cnt
 
# Given array
arr = [ 6, 4, 2, 3 ]
n = len(arr)
 
# Function Call
print(count(arr, n))
 
# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;
class GFG{
 
// Method to check if x is power of 2
static bool check(int x)
{
     
    // First x in the below expression
    // is for the case when x is 0
    return x != 0 && ((x & (x - 1)) == 0);
}
 
// Function to return the
// number of valid pairs
static int count(int []arr, int n)
{
    int cnt = 0;
 
    // Iterate for all possible pairs
    for(int i = 0; i < n - 1; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
            
          // Bitwise and value of
          // the pair is passed
          if (check(arr[i] & arr[j]))
              cnt++;
       }
    }
     
    // Return the final count
    return cnt;
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = new int[]{ 6, 4, 2, 3 };
 
    int n = arr.Length;
     
    // Function call
    Console.Write(count(arr, n));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// JavaScript program to implement
// the above approach
 
// Method to check if x is power of 2
function check(x)
{
   
    // First x in the below expression
    // is for the case when x is 0
    return x != 0 && ((x & (x - 1)) == 0);
}
   
// Function to return the
// number of valid pairs
function count(arr, n)
{
    let cnt = 0;
   
    // Iterate for all possible pairs
    for(let i = 0; i < n - 1; i++)
    {
       for(let j = i + 1; j < n; j++)
       {
             
          // Bitwise and value of
          // the pair is passed
          if (check(arr[i] & arr[j]))
              cnt++;
       }
    }
       
    // Return the final count
    return cnt;
}
 
// Driver code
    // Given array arr[]
    let arr = [ 6, 4, 2, 3 ];
   
    let n = arr.length;
       
    // Function call
    document.write(count(arr, n));
 
// This code is contributed by susmitakundugoaldanga.

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live