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Count ordered pairs of Array elements such that bitwise AND of K and XOR of the pair is 0

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  • Difficulty Level : Basic
  • Last Updated : 29 Jul, 2022

Given an array arr[] of size N and an integer K, the task is to find the count of all the ordered pairs (i, j) where i != j, such that ((arr[i] ⊕ arr[j]) & K) = 0. The represents bitwise XOR and & represents bitwise AND.

Examples:

Input: arr = [1, 2, 3, 4, 5], K = 3
Output: 2
Explanation: There are 2 pairs satisfying the condition. 
These pairs are: (1, 5) and (5, 1)

Input: arr = [5, 9, 24], K = 7
Output: 0
Explanation: No such pair satisfying the condition exists.

 

Approach: The given problem can be solved with the help of the following idea:

Using distributive property, we can write ((arr[i] ⊕ arr[j]) & K) = ((arr[i] & K) ⊕ (arr[j] & K))
Since for ((arr[i] & K) ⊕ (arr[j] & K)) = 0, these two terms (arr[i] & K) and (arr[j] & K) must be equal.

Follow the below steps to solve the problem:

  • Create a map and an answer variable (say Res = 0).
  • Traverse the array and insert (arr[i] & K) to map with its count.
  • Now, traverse the map and for each entry if there are X such occurrences then possible pairs = X*(X-1). So add that to the value Res.
  • Return Res as the required answer.

Below is the implementation of the above approach:

C++




// C++ code to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find pair satisfying the condition
int findPair(int* arr, int N, int K)
{
    map<int, int> Mp;
    int Res = 0;
    for (int i = 0; i < N; i++)
        Mp[arr[i] & K]++;
 
    for (auto i : Mp)
        Res += ((i.second - 1) * i.second);
 
    return Res;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    // Function call
    cout << findPair(arr, N, K) << endl;
    return 0;
}

Java




// Java code to implement the above approach
import java.util.*;
 
class GFG {
 
  // Function to find pair satisfying the condition
  static int findPair(int arr[], int N, int K)
  {
    Map<Integer, Integer> mp = new HashMap<>();
    int Res = 0;
 
    for (int i = 0; i < N; i++)
    {
      if (mp.containsKey((arr[i] & K)))
      {
        mp.put((arr[i] & K), mp.get((arr[i] & K)) + 1);
      }
      else
      {
        mp.put((arr[i] & K), 1);
      }
    }
    for (Map.Entry<Integer, Integer> i : mp.entrySet())
    {
      Res += ((i.getValue() - 1) * i.getValue());
    }
 
    return Res;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = arr.length;
    int K = 3;
 
    // Function call
    System.out.println(findPair(arr, N, K));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3




# Python3 code to implement the above approach
 
# Function to find pair satisfying the condition
def findPair(arr, N, K) :
    Mp = {};
    Res = 0;
 
    for i in range(N) :
        Mp[arr[i] & K] = Mp.get(arr[i] & K, 0) + 1;
 
    for i in Mp:
        Res += ((Mp[i] - 1) * Mp[i]);
 
    return Res;
 
 
# Driver Code
if __name__ == "__main__" :
    arr = [ 1, 2, 3, 4, 5 ];
    N = len(arr);
    K = 3;
 
    # Function call
    print(findPair(arr, N, K));
     
    # This code is contributed by AnkThon

C#




using System;
using System.Collections.Generic;
public class GFG{
   
  // Function to find pair satisfying the condition
public static int findPair(int[] arr, int N, int K)
{
    SortedDictionary<int,int>Mp=
            new SortedDictionary<int,int>();
    int Res = 0;
   
  // Initialising Mp with 0
  for(int i=0;i<N;i++)
  {
    Mp[i] = 0;
  }
 
    for (int i = 0; i < N; i++)
        Mp[(arr[i] & K)]++;
 
   
     foreach( KeyValuePair<int,int> kvp in Mp )
        {
            Res+=(( kvp.Value - 1) *  kvp.Value);
        }
 
    return Res;
}
 
 
    static public void Main (){
 
        int[] arr = { 1, 2, 3, 4, 5 };
    int N =arr.Length;
    int K = 3;
 
    // Function call
    Console.WriteLine( findPair(arr, N, K) );
    }
}
 
// This code is contributed by akashish__

Javascript




<script>
        // JavaScript code for the above approach
 
        // Function to find pair satisfying the condition
        function findPair(arr, N, K) {
            let Mp = new Map();
            let Res = 0;
 
            for (let i = 0; i < N; i++) {
                Mp[arr[i] & K]++;
 
                if (Mp.has(arr[i] & K)) {
                    Mp.set(arr[i] & K, Mp.get(arr[i] & K) + 1)
                }
                else {
                    Mp.set(arr[i] & K, 1);
                }
            }
            for (let [key, val] of Mp)
                Res += ((val - 1) * val);
 
            return Res;
        }
 
        // Driver Code
 
        let arr = [1, 2, 3, 4, 5];
        let N = arr.length;
        let K = 3;
 
        // Function call
        document.write(findPair(arr, N, K) + '<br>');
 
    // This code is contributed by Potta Lokesh
    </script>

Output

2

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


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