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Maximize count of pairs whose Bitwise AND exceeds Bitwise XOR by replacing such pairs with their Bitwise AND
  • Last Updated : 26 Feb, 2021

Given an array arr[] consisting of N positive integers, replace pairs of array elements whose Bitwise AND exceeds Bitwise XOR values by their Bitwise AND value. Finally, count the maximum number of such pairs that can be generated from the array.

Examples:

Input: arr[] = {12, 9, 15, 7}
Output: 2
Explanation:
Step 1: Select the pair {12, 15} and replace the pair by their Bitwise AND (= 12). The array arr[] modifies to {12, 9, 7}. 
Step 2: Replace the pair {12, 9} by their Bitwise AND (= 8). Therefore, the array arr[] modifies to {8, 7}. 
Therefore, the maximum number of such pairs is 2.

Input: arr[] = {2, 6, 12, 18, 9}
Output: 1

Naive Approach: The simplest approach to solve this problem is to generate all possible pairs and select a pair having Bitwise AND greater than their Bitwise XOR . Replace the pair and insert their Bitwise AND. Repeat the above process until no such pairs are found. Print the count of such pairs obtained. 
Time Complexity: O(N3)
Auxiliary Space: O(1)



Efficient Approach: The above approach can be optimized based on the following observations:

  • A number having its most significant bit at the ith position can only form a pair with other numbers having MSB at the ith position.
  • The total count of numbers having their MSB at the ith position decreases by one if one of these pairs is selected.
  • Thus, the total pairs that can be formed at the ith position are the total count of numbers having MB at ith position decreased by 1.

Follow the steps below to solve the problem:

  • Initialize a Map, say freq, to store the count of numbers having MSB at respective bit positions.
  • Traverse the array and for each array element arr[i], find the MSB of arr[i] and increment the count of MSB in the freq by 1.
  • Initialize a variable, say pairs, to store the count of total pairs.
  • Traverse the map and update pairs as pairs += (freq[i] – 1).
  • After completing the above steps, print the value of pairs 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 whose Bitwise AND is
// greater than the Bitwise XOR
int countPairs(int arr[], int N)
{
    // Stores the frequency of
    // MSB of array elements
    unordered_map<int, int> freq;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Increment count of numbers
        // having MSB at log(arr[i])
        freq[log2(arr[i])]++;
    }
 
    // Stores total number of pairs
    int pairs = 0;
 
    // Traverse the Map
    for (auto i : freq) {
        pairs += i.second - 1;
    }
 
    // Return total count of pairs
    return pairs;
}
 
// Driver Code
int main()
{
    int arr[] = { 12, 9, 15, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, N);
 
    return 0;
}


Python3




# Python3 program for the above approach
from math import log2
 
# Function to count the number of
# pairs whose Bitwise AND is
# greater than the Bitwise XOR
def countPairs(arr, N):
   
    # Stores the frequency of
    # MSB of array elements
    freq = {}
 
    # Traverse the array
    for i in range(N):
 
        # Increment count of numbers
        # having MSB at log(arr[i])
        x = int(log2(arr[i]))
        freq[x] = freq.get(x, 0) + 1
 
    # Stores total number of pairs
    pairs = 0
 
    # Traverse the Map
    for i in freq:
        pairs += freq[i] - 1
 
    # Return total count of pairs
    return pairs
 
# Driver Code
if __name__ == '__main__':
    arr = [12, 9, 15, 7]
    N = len(arr)
    print(countPairs(arr, N))
 
    # This code is contributed by mohit kumar 29.


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
// Function to count the number of
// pairs whose Bitwise AND is
// greater than the Bitwise XOR
static int countPairs(int []arr, int N)
{
   
    // Stores the frequency of
    // MSB of array elements
    Dictionary<int,int> freq = new Dictionary<int,int>();
 
    // Traverse the array
    for (int i = 0; i < N; i++)
    {
 
        // Increment count of numbers
        // having MSB at log(arr[i])
        if(freq.ContainsKey((int)(Math.Log(Convert.ToDouble(arr[i]),2.0))))
         freq[(int)(Math.Log(Convert.ToDouble(arr[i]),2.0))]++;
        else
          freq[(int)(Math.Log(Convert.ToDouble(arr[i]),2.0))] = 1;
    }
 
    // Stores total number of pairs
    int pairs = 0;
 
    // Traverse the Map
    foreach(var item in freq)
    {
        pairs += item.Value - 1;
    }
 
    // Return total count of pairs
    return pairs;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 12, 9, 15, 7 };
    int N =  arr.Length;
    Console.WriteLine(countPairs(arr, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.


Output: 

2

 

Time Complexity: O(N)
Auxiliary Space: O(32)

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