Skip to content
Related Articles

Related Articles

Improve Article

Count pairs from an array whose Bitwise OR is greater than Bitwise AND

  • Last Updated : 04 May, 2021

Given an array A[] consisting of N integers, the task is to count the number of pairs (i, j) such that i < j, and Bitwise OR of A[i] and A[j] is greater than Bitwise AND of A[i] and A[j].

Examples:

Input: A[] = {1, 4, 7}
Output: 3
Explanation: 
There are 3 such pairs: (1, 4), (4, 7), (1, 7).
1) 1 | 4 (= 5) > 1 & 4 (= 0)
2) 4 | 7 (= 7) > 4 & 7 (= 4)
3) 1 | 7 (= 7) > 1 & 7 (= 1)

Input: A[] = {3, 3, 3}
Output: 0
Explanation: There exist no such pair.

Naive Approach: The simplest approach is to generate all possible pairs from the given array and for every pair (i, j), if it satisfies the given conditions, then increment the count by 1. After checking for all the pairs, print the value of count as the result.



Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// pairs (i, j) their Bitwise OR is
// greater than Bitwise AND
void countPairs(int A[], int n)
{
    // Store the required answer
    int count = 0;
 
    // Check for all possible pairs
    for (int i = 0; i < n; i++) {
 
        for (int j = i + 1; j < n; j++)
 
            // If the condition satisfy
            // then increment count by 1
            if ((A[i] | A[j])
                > (A[i] & A[j])) {
                count++;
            }
    }
 
    // Print the answer
    cout << count;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 4, 7 };
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    countPairs(A, N);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
  // Function to count the number of
  // pairs (i, j) their Bitwise OR is
  // greater than Bitwise AND
  static void countPairs(int A[], int n)
  {
    // Store the required answer
    int count = 0;
 
    // Check for all possible pairs
    for (int i = 0; i < n; i++) {
 
      for (int j = i + 1; j < n; j++)
 
        // If the condition satisfy
        // then increment count by 1
        if ((A[i] | A[j])
            > (A[i] & A[j])) {
          count++;
        }
    }
 
    // Print the answer
    System.out.println(count);
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int A[] = { 1, 4, 7 };
    int N = A.length;
 
    // Function Call
    countPairs(A, N);
  }
}
 
// This code is contributed by splevel62.

Python3




# Python program to implement
# the above approach
 
# Function to count the number of
# pairs (i, j) their Bitwise OR is
# greater than Bitwise AND
def countPairs(A, n):
   
    # Store the required answer
    count = 0;
 
    # Check for all possible pairs
    for i in range(n):
        for j in range(i + 1, n):
 
            # If the condition satisfy
            # then increment count by 1
            if ((A[i] | A[j]) > (A[i] & A[j])):
                count += 1;
 
    # Prthe answer
    print(count);
 
# Driver Code
if __name__ == '__main__':
    A = [1, 4, 7];
    N = len(A);
 
    # Function Call
    countPairs(A, N);
 
    # This code is contributed by 29AjayKumar

C#




// C# program to implement
// the above approach
using System;
class GFG
{
  // Function to count the number of
  // pairs (i, j) their Bitwise OR is
  // greater than Bitwise AND
  static void countPairs(int[] A, int n)
  {
    // Store the required answer
    int count = 0;
 
    // Check for all possible pairs
    for (int i = 0; i < n; i++) {
 
      for (int j = i + 1; j < n; j++)
 
        // If the condition satisfy
        // then increment count by 1
        if ((A[i] | A[j])
            > (A[i] & A[j])) {
          count++;
        }
    }
 
    // Print the answer
    Console.Write(count);
}
 
// Driver Code
public static void Main()
{
    int[] A = { 1, 4, 7 };
    int N = A.Length;
 
    // Function Call
    countPairs(A, N);
}
}
 
// This code is contributed by susmitakundugoaldanga.

Javascript




<script>
 
 
// Javascript program for the above approach
 
// Function to count the number of
// pairs (i, j) their Bitwise OR is
// greater than Bitwise AND
function countPairs( A, n)
{
    // Store the required answer
    var count = 0;
 
    // Check for all possible pairs
    for (var i = 0; i < n; i++) {
 
        for (var j = i + 1; j < n; j++)
 
            // If the condition satisfy
            // then increment count by 1
            if ((A[i] | A[j])
                > (A[i] & A[j])) {
                count++;
            }
    }
 
    // Print the answer
    document.write( count);
}
 
// Driver Code
var A = [ 1, 4, 7 ];
var N = A.length;
// Function Call
countPairs(A, N);
 
</script>
Output: 
3

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the observation that the condition Ai | Aj > Ai & Aj is satisfied only when Ai and Aj are distinct. If Ai is equal to Aj, then Ai | Aj = Ai & Aj. The condition Ai | Aj < Ai & Aj is never met. Hence, the problem can be solved by subtracting the number of pairs having an equal value from the total number of pairs possible. Follow the steps below to solve the problem:

  • Initialize a variable, say count, with N * (N – 1) / 2 that denotes the total number of possible pairs.
  • Initialize a hashmap, say M to store the frequency of each array element.
  • Traverse the array arr[] and for each array element arr[i], increment the frequency of arr[i] in M by 1.
  • Now, traverse the hashmap M and for each key K, and its corresponding value X, subtract the value of (X * (X – 1))/2 from the count.
  • After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// pairs (i, j) their Bitwise OR is
// greater than Bitwise AND
void countPairs(int A[], int n)
{
    // Total number of pairs
    // possible from the array
    long long count = (n * (n - 1)) / 2;
 
    // Stores frequency of each array element
    unordered_map<long long, long long> ump;
 
    // Traverse the array A[]
    for (int i = 0; i < n; i++) {
 
        // Increment ump[A[i]] by 1
        ump[A[i]]++;
    }
 
    // Traverse the Hashmap ump
    for (auto it = ump.begin();
         it != ump.end(); ++it) {
 
        // Subtract those pairs (i, j)
        // from count which has the
        // same element on index i
        // and j (i < j)
        long long c = it->second;
        count = count - (c * (c - 1)) / 2;
    }
 
    // Print the result
    cout << count;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 4, 7 };
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    countPairs(A, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to count the number of
// pairs (i, j) their Bitwise OR is
// greater than Bitwise AND
static void countPairs(int A[], int n)
{
    // Total number of pairs
    // possible from the array
    int count = (n * (n - 1)) / 2;
 
    // Stores frequency of each array element
    Map<Integer,Integer> mp = new HashMap<>();
 
    // Traverse the array A[]
    for (int i = 0 ; i < n; i++){
        if(mp.containsKey(A[i])){
            mp.put(A[i], mp.get(A[i])+1);
        }
        else{
            mp.put(A[i], 1);
        }
    }
 
    // Traverse the Hashmap ump
    for (Map.Entry<Integer,Integer> it : mp.entrySet()){
 
        // Subtract those pairs (i, j)
        // from count which has the
        // same element on index i
        // and j (i < j)
        int c = it.getValue();
        count = count - (c * (c - 1)) / 2;
    }
 
    // Print the result
    System.out.print(count);
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 1, 4, 7 };
    int N = A.length;
 
    // Function Call
    countPairs(A, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python 3 program for the above approach
from collections import defaultdict
 
# Function to count the number of
# pairs (i, j) their Bitwise OR is
# greater than Bitwise AND
def countPairs(A, n):
 
    # Total number of pairs
    # possible from the array
    count = (n * (n - 1)) // 2
 
    # Stores frequency of each array element
    ump = defaultdict(int)
 
    # Traverse the array A[]
    for i in range(n):
 
        # Increment ump[A[i]] by 1
        ump[A[i]] += 1
 
    # Traverse the Hashmap ump
    for it in ump.keys():
 
        # Subtract those pairs (i, j)
        # from count which has the
        # same element on index i
        # and j (i < j)
        c = ump[it]
        count = count - (c * (c - 1)) // 2
 
    # Print the result
    print(count)
 
# Driver Code
if __name__ == "__main__":
 
    A = [1, 4, 7]
    N = len(A)
 
    # Function Call
    countPairs(A, N)
 
    # This code is contributed by chitranayal.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to count the number of
  // pairs (i, j) their Bitwise OR is
  // greater than Bitwise AND
  static void countPairs(int[] A, int n)
  {
 
    // Total number of pairs
    // possible from the array
    long count = (n * (n - 1)) / 2;
 
    // Stores frequency of each array element
    Dictionary<long, long> ump = new Dictionary<long, long>();
 
    // Traverse the array A[]
    for (int i = 0; i < n; i++)
    {
 
      // Increment ump[A[i]] by 1
      if(ump.ContainsKey(A[i]))
      {
        ump[A[i]]++;
      }
      else{
        ump[A[i]] = 1;
      }
    }
 
    // Traverse the Hashmap ump
    foreach(KeyValuePair<long, long> it in ump)
    {
 
      // Subtract those pairs (i, j)
      // from count which has the
      // same element on index i
      // and j (i < j)
      long c = it.Value;
      count = count - (c * (c - 1)) / 2;
    }
 
    // Print the result
    Console.WriteLine(count);
  }
 
  // Driver code
  static void Main() {
    int[] A = { 1, 4, 7 };
    int N = A.Length;
 
    // Function Call
    countPairs(A, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.
Output: 
3

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :