Count pairs with equal Bitwise AND and Bitwise OR value

Given an array, arr[] of size N, the task is to count the number of unordered pairs such that Bitwise AND and Bitwise OR of each pair is equal.

Examples:

Input: arr[] = {1, 2, 1} 
Output:
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

Naive Approach: The simplest approach to solve the problem is to traverse the array and generate all possible pairs of the given array. For each pair, check if Bitwise And of the pair is equal to Bitwise OR of that pair or not. If found to be true, then increment the counter. Finally, print the value of the counter.



Efficient Approach: To optimize the above approach the idea is based on the following observations:

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. 

Follow the steps below to solve the problem:

  • Initialize a variable, say cntPairs to store the count of pairs whose Bitwise AND(&) value and Bitwise OR(|) value is equal.
  • Create a map, say mp to store the frequency of all distinct elements of the given array.
  • Traverse the given array and store the frequency of all distinct elements of the given array in mp.
  • Traverse map and check if frequency, say freq is greater than 1 then update cntPairs += (freq * (freq – 1)) / 2.
  • Finally, print the value of cntPairs.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
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<int, int> 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<<countPairs(arr, N);
}

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Java

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// 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<Integer, Integer> 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<Integer, Integer> 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

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Python3

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# 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

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C#

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// 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<int,
               int> mp = new Dictionary<int,
                                        int>();
 
    // 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<int, int> 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

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Output: 

4




 

Time Complexity: O(N) 
Auxiliary Space: O(N)

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Improved By : akhilsaini