# Count pairs with equal Bitwise AND and Bitwise OR value

• Difficulty Level : Medium
• Last Updated : 20 May, 2021

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, arr) is (arr & arr) = (1 & 2) = 0
Bitwise AND of the pair(arr, arr) is (arr | arr) = (1 | 2) = 3
Bitwise AND of the pair(arr, arr) is (arr & arr) = (1 & 2) = 1
Bitwise AND of the pair(arr, arr) is (arr | arr) = (1 | 2) = 1
Bitwise AND of the pair(arr, arr) is (arr & arr) = (2 & 1) = 0
Bitwise AND of the pair(arr, arr) is arr | arr = (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++

 `// 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<``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);` `    ``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<``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`

## Javascript

 ``

Output:

`4`

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

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