Count pairs with equal Bitwise AND and Bitwise OR value - GeeksforGeeks

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: 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
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++
Java
Python3
C#
Javascript
// 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 ORint 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 Codeint main(){    int arr[] = { 1, 2, 3, 1, 2, 2 };    int N = sizeof(arr) / sizeof(arr[0]);    cout<<countPairs(arr, N);}
// Java program to implement// the above approachimport java.io.*;import java.util.*; class GFG{ // Function to count pairs in an array// whose bitwise AND equal to bitwise ORstatic 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 Codepublic 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 program to implement# the above approach # Function to count pairs in an array# whose bitwise AND equal to bitwise ORdef 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 Codeif __name__ == "__main__":     arr = [ 1, 2, 3, 1, 2, 2 ]    N = len(arr)         print(countPairs(arr, N)) # This code is contributed by akhilsaini
// C# program to implement// the above approachusing System;using System.Collections.Generic; class GFG{ // Function to count pairs in an array// whose bitwise AND equal to bitwise ORstatic 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 Codepublic 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
<script>  // JavaScript program to implement// the above approach // Function to count pairs in an array// whose bitwise AND equal to bitwise ORfunction 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 Codevar arr = [1, 2, 3, 1, 2, 2 ];var N = arr.length;document.write( countPairs(arr, N)); </script>
4
 
Time Complexity: O(N) Auxiliary Space: O(N)
akhilsaini
rrrtnx
Bitwise-AND
Bitwise-OR
cpp-map
frequency-counting
Arrays
Bit Magic
Greedy
Mathematical
Pattern Searching
Arrays
Greedy
Mathematical
Bit Magic
Pattern Searching
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