对具有相等的按位 AND 和按位 OR 值的对进行计数

给定一个大小为N的数组arr[] ，任务是计算无序对的数量，使得每对的按位 AND 和按位 OR 相等。

例子：

Input: arr[] = {1, 2, 1} 
Output: 1 
Explanation: 
Bitwise AND value and Bitwise OR value all possible pairs are: 
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] & arr[1]) = (1 & 2) = 0 
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] | arr[1]) = (1 | 2) = 3 
Bitwise AND of the pair(arr[0], arr[2]) is (arr[0] & arr[2]) = (1 & 2) = 1 
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] | arr[1]) = (1 | 2) = 1 
Bitwise AND of the pair(arr[1], arr[2]) is (arr[1] & arr[2]) = (2 & 1) = 0 
Bitwise AND of the pair(arr[0], arr[1]) is arr[0] | arr[1] = (2 | 1) = 3 
Therefore, the required output is 1. 

Input: arr[] = {1, 2, 3, 1, 2, 2} 
Output: 4

朴素方法：解决问题的最简单方法是遍历数组并生成给定数组的所有可能对。对于每一对，检查该对的 Bitwise And 是否等于该对的 Bitwise OR。如果发现为真，则增加计数器。最后，打印计数器的值。

高效的方法：为了优化上述方法，该想法基于以下观察：

0 & 0 = 0 and 0 | 0 = 0 
0 & 1 = 0 and 0 | 1 = 1 
1 & 0 = 0 and 1 | 0 = 1 
1 & 1 = 1 and 1 | 1 = 1 
 

Therefore, If both the elements of a pair are equal, only then, bitwise AND(&) and Bitwise OR(|) of the pair becomes equal. 

请按照以下步骤解决问题：

• 初始化一个变量，比如cntPairs来存储其按位 AND(&) 值和按位 OR(|) 值相等的对的计数。
• 创建一个映射，比如mp来存储给定数组的所有不同元素的频率。
• 遍历给定数组并将给定数组的所有不同元素的频率存储在mp 中。
• 遍历地图并检查频率是否大于1，然后更新cntPairs += (freq * (freq – 1)) / 2。
• 最后，打印cntPairs的值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to count pairs in an array
// whose bitwise AND equal to bitwise OR
int countPairs(int arr[], int N)
{
     
    // Store count of pairs whose
    // bitwise AND equal to bitwise OR
    int cntPairs = 0;
     
    // Stores frequency of
    // distinct elements of array
    map mp;
     
    // Traverse the array
    for (int i = 0; i < N; i++) {
         
        // Increment the frequency
        // of arr[i]
        mp[arr[i]]++;
    }
     
    // Traverse map
    for (auto freq: mp) {
        cntPairs += (freq.second *
                   (freq.second - 1)) / 2;
    }
     
    return cntPairs;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 1, 2, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout<

---

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to count pairs in an array
// whose bitwise AND equal to bitwise OR
static int countPairs(int[] arr, int N)
{
     
    // Store count of pairs whose
    // bitwise AND equal to bitwise OR
    int cntPairs = 0;
 
    // Stores frequency of
    // distinct elements of array
    HashMap mp = new HashMap<>();
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Increment the frequency
        // of arr[i]
        mp.put(arr[i],
               mp.getOrDefault(arr[i], 0) + 1);
    }
 
    // Traverse map
    for(Map.Entry freq : mp.entrySet())
    {
        cntPairs += (freq.getValue() *
                    (freq.getValue() - 1)) / 2;
    }
 
    return cntPairs;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 1, 2, 2 };
    int N = arr.length;
     
    System.out.println(countPairs(arr, N));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
 
# Function to count pairs in an array
# whose bitwise AND equal to bitwise OR
def countPairs(arr, N):
     
    # Store count of pairs whose
    # bitwise AND equal to bitwise OR
    cntPairs = 0
 
    # Stores frequency of
    # distinct elements of array
    mp = {}
 
    # Traverse the array
    for i in range(0, N):
         
        # Increment the frequency
        # of arr[i]
        if arr[i] in mp:
            mp[arr[i]] = mp[arr[i]] + 1
        else:
            mp[arr[i]] = 1
 
    # Traverse map
    for freq in mp:
        cntPairs += int((mp[freq] *
                        (mp[freq] - 1)) / 2)
 
    return cntPairs
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 1, 2, 3, 1, 2, 2 ]
    N = len(arr)
     
    print(countPairs(arr, N))
 
# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count pairs in an array
// whose bitwise AND equal to bitwise OR
static int countPairs(int[] arr, int N)
{
     
    // Store count of pairs whose
    // bitwise AND equal to bitwise OR
    int cntPairs = 0;
 
    // Stores frequency of
    // distinct elements of array
    Dictionary mp = new Dictionary();
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Increment the frequency
        // of arr[i]
        if (!mp.ContainsKey(arr[i]))
            mp.Add(arr[i], 1);
        else
            mp[arr[i]] = mp[arr[i]] + 1;
    }
 
    // Traverse map
    foreach(KeyValuePair freq in mp)
    {
        cntPairs += (freq.Value *
                    (freq.Value - 1)) / 2;
    }
 
    return cntPairs;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 2, 3, 1, 2, 2 };
    int N = arr.Length;
     
    Console.WriteLine(countPairs(arr, N));
}
}
 
// This code is contributed by akhilsaini

Javascript

  
// JavaScript program to implement
// the above approach
 
// Function to count pairs in an array
// whose bitwise AND equal to bitwise OR
function countPairs(arr, N)
{
     
    // Store count of pairs whose
    // bitwise AND equal to bitwise OR
    var cntPairs = 0;
     
    // Stores frequency of
    // distinct elements of array
    var mp = new Map();
     
    // Traverse the array
    for (var i = 0; i < N; i++) {
         
        // Increment the frequency
        // of arr[i]
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1);
    }
     
    // Traverse map
    mp.forEach((value, key) => {
         
        cntPairs += parseInt((value *
                   (value - 1)) / 2);
    });
     
    return cntPairs;
}
 
// Driver Code
var arr = [1, 2, 3, 1, 2, 2 ];
var N = arr.length;
document.write( countPairs(arr, N));
 

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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