按位与为2的幂的对的计数

给定N个正整数的数组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的对，检查每对元素的按位与的想法是否为2的乘方幂。如果“是”，则计算该对，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

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)