Count pairs with bitwise XOR exceeding bitwise AND from a given array

Given an array, arr[] of size N, the task is to count the number of pairs from the given array such that the bitwise AND(&) of each pair is less than its bitwise XOR(^).

Examples:

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 8
Explanation:
Pairs that satisfy the given conditions are: 
(1 & 2) < (1 ^ 2)
(1 & 3) < (1 ^ 3)
(1 & 4) < (1 ^ 4)
(1 & 5) < (1 ^ 5)
(2 & 4) < (2 ^ 4)
(2 & 5) < (2 ^ 5)
(3 & 4) < (3 ^ 4)
(3 & 5) < (3 ^ 5)
Therefore, the required output is 8. 

Input: arr[] = {1, 4, 3, 7, 10} 
Output: 9 

Approach: The simplest approach is to traverse the array and generate all possible pairs from the given array. For each pair, check if its bitwise AND(&) is less than the bitwise XOR(^) of that pair or not. If found to be true, then increment the count of pairs by 1. Finally, print the count of such pairs obtained. 

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

Efficient approach: To optimize the above approach, follow the properties of the bitwise operators:

1 ^ 0 = 1
0 ^ 1 = 1
1 & 1 = 1
X = b₃₁b₃₀…..b₁b₀
Y = a₃₁b₃₀….a₁a₀
If the Expression {(X & Y) > (X ^ Y)} is true then the most significant bit(MSB) of both X and Y must be equal. 
Total count of pairs that satisfy the condition{(X & Y) > (X ^ Y)} are: 

 bit[i] stores the count of array elements whose position of most significant bit(MSB) is i.

Therefore, total count of pairs that satisfy the given condition{(X & Y) < (X ^ Y)} 
= [{N * (N – 1) /2} – {

 }]

Follow the steps below to solve the problem: 

1. Initialize a variable, say res, to store the count of pairs that satisfy the given condition.
2. Traverse the given array.
3. Store the position of most significant bit of each element of the given array.
4. Finally, evaluate the result by the above mentioned formula and print the result.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to count pairs that
// satisfy the above condition.
int cntPairs(int arr[], int N)
{

    // Stores the count
    // of pairs
    int res = 0;

    // Stores the count of array
    // elements having same
    // positions of MSB
    int bit[32] = { 0 };

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Stores the index of
        // MSB of array elements
        int pos
            = log2(arr[i]);
        bit[pos]++;
    }

    // Calculate number of pairs
    for (int i = 0; i < 32; i++) {
        res += (bit[i]
                * (bit[i] - 1))
               / 2;
    }
    res = (N * (N - 1)) / 2 - res;

    return res;
}

// Driver Code
int main()
{

    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << cntPairs(arr, N);
}

Java

// Java program to implement
// the above approach
import java.io.*;

class GFG{

// Function to count pairs that
// satisfy the above condition.
static int cntPairs(int[] arr, int N)
{

    // Stores the count
    // of pairs
    int res = 0;

    // Stores the count of array
    // elements having same
    // positions of MSB
    int[] bit = new int[32];

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Stores the index of
        // MSB of array elements
        int pos = (int)(Math.log(arr[i]) /
                        Math.log(2));
        bit[pos]++;
    }

    // Calculate number of pairs
    for(int i = 0; i < 32; i++)
    {
        res += (bit[i] * (bit[i] - 1)) / 2;
    }
    res = (N * (N - 1)) / 2 - res;

    return res;
}

// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int N = arr.length;

    System.out.println(cntPairs(arr, N));
}
}

// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
import math

# Function to count pairs that
# satisfy the above condition.
def cntPairs(arr, N):

    # Stores the count
    # of pairs
    res = 0

    # Stores the count of array
    # elements having same
    # positions of MSB
    bit = [0] * 32

    # Traverse the array
    for i in range(0, N):

        # Stores the index of
        # MSB of array elements
        pos = int(math.log(arr[i], 2))
        bit[pos] = bit[pos] + 1

    # Calculate number of pairs
    for i in range(0, 32):
        res = res + int((bit[i] *
                        (bit[i] - 1)) / 2)

    res = int((N * (N - 1)) / 2 - res)

    return res

# Driver Code
if __name__ == "__main__":

    arr = [ 1, 2, 3, 4, 5, 6 ]
    N = len(arr)

    print(cntPairs(arr, N))

# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to count pairs that
// satisfy the above condition.
static int cntPairs(int[] arr, int N)
{

    // Stores the count
    // of pairs
    int res = 0;

    // Stores the count of array
    // elements having same
    // positions of MSB
    int[] bit = new int[32];

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // Stores the index of
        // MSB of array elements
        int pos = (int)(Math.Log(arr[i]) /
                        Math.Log(2));
        bit[pos]++;
    }

    // Calculate number of pairs
    for(int i = 0; i < 32; i++)
    {
        res += (bit[i] * (bit[i] - 1)) / 2;
    }
    res = (N * (N - 1)) / 2 - res;

    return res;
}

// Driver Code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int N = arr.Length;

    Console.Write(cntPairs(arr, N));
}
}

// This code is contributed by akhilsaini

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count pairs that
// satisfy the above condition.
function cntPairs(arr, N)
{

    // Stores the count
    // of pairs
    let res = 0;

    // Stores the count of array
    // elements having same
    // positions of MSB
    let bit = new Array(32).fill(0);

    // Traverse the array
    for(let i = 0; i < N; i++)
    {

        // Stores the index of
        // MSB of array elements
        let pos = parseInt(Math.log(arr[i]) /
                           Math.log(2));;
        bit[pos]++;
    }

    // Calculate number of pairs
    for(let i = 0; i < 32; i++)
    {
        res += parseInt((bit[i]
                * (bit[i] - 1)) / 2);
    }
    res = parseInt((N * (N - 1)) / 2) - res;

    return res;
}

// Driver Code
let arr = [ 1, 2, 3, 4, 5, 6 ];
let N = arr.length;

document.write(cntPairs(arr, N));

// This code is contributed by subhammahato348.

</script>

Output:

11

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

Method 2 : Bitwise and is greater than bitwise xor if and only if most significant bit is equal.

• Create a bits[] array of size 32 (max no of bits)
• Initialize ans to 0.
• We will traverse the array from the start and for each number,
  • Find its most significant bit and say it is j.
  • Add the value stored in bits[j] array to the ans. (for the current element bits[j] number of pairs can be formed)
  • Now increase the value of bits[j] by 1.
• Now total number of pairs = n*(n-1)/2. Subtract the ans from it.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

int findCount(int arr[], int N)
{
    // For storing number of pairs
    int ans = 0;

    // For storing count of numbers
    int bits[32] = { 0 };

    // Iterate from 0 to N - 1
    for (int i = 0; i < N; i++) {

        // Find the most significant bit
        int val = log2l(arr[i]);

        ans += bits[val];
        bits[val]++;
    }
    return N * (N - 1) / 2 - ans;
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4, 5, 6 };

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << findCount(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG{

static int findCount(int arr[], int N)
{

    // For storing number of pairs
    int ans = 0;

    // For storing count of numbers
    int bits[] = new int[32];

    // Iterate from 0 to N - 1
    for(int i = 0; i < N; i++)
    {

        // Find the most significant bit
        int val = (int)(Math.log(arr[i]) /
                        Math.log(2));

        ans += bits[val];
        bits[val]++;
    }
    return N * (N - 1) / 2 - ans;
}

// Driver Code
public static void main(String[] args)
{

    // Given array arr[]
    int arr[] = { 1, 2, 3, 4, 5, 6 };

    int N = arr.length;

    // Function Call
    System.out.println(findCount(arr, N));
}
}

// This code is contributed by Kingash

Python3

# Python3 program for the above approach
import math
def findCount(arr, N):

    # For storing number of pairs
    ans = 0

    # For storing count of numbers
    bits = [0] * 32

    # Iterate from 0 to N - 1
    for i in range(N):

        # Find the most significant bit
        val = int(math.log2(arr[i]))

        ans += bits[val]
        bits[val] += 1
    return (N * (N - 1) // 2 - ans)

# Driver Code
if __name__ == "__main__":

    # Given array arr[]
    arr = [1, 2, 3, 4, 5, 6]

    N = len(arr)

    # Function Call
    print(findCount(arr, N))

    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;

class GFG{

static int findCount(int[] arr, int N)
{

    // For storing number of pairs
    int ans = 0;

    // For storing count of numbers
    int[] bits = new int[32];

    // Iterate from 0 to N - 1
    for(int i = 0; i < N; i++)
    {

        // Find the most significant bit
        int val = (int)(Math.Log(arr[i]) /
                        Math.Log(2));

        ans += bits[val];
        bits[val]++;
    }
    return N * (N - 1) / 2 - ans;
}

// Driver Code
public static void Main()
{

    // Given array arr[]
    int[] arr = { 1, 2, 3, 4, 5, 6 };

    int N = arr.Length;

    // Function Call
    Console.Write(findCount(arr, N));
}
}

// This code is contributed by subhammahato348

Javascript

<script>

// Javascript program for the above approach
function findCount(arr, N)
{

    // For storing number of pairs
    let ans = 0;

    // For storing count of numbers
    let bits = new Array(32).fill(0);

    // Iterate from 0 to N - 1
    for(let i = 0; i < N; i++)
    {

        // Find the most significant bit
        let val = parseInt(Math.log(arr[i]) /
                           Math.log(2));

        ans += bits[val];
        bits[val]++;
    }
    return parseInt(N * (N - 1) / 2) - ans;
}

// Driver Code

// Given array arr[]
let arr = [ 1, 2, 3, 4, 5, 6 ];

let N = arr.length;

// Function Call
document.write(findCount(arr, N));

// This code is contributed by subhammahato348

</script>

Output

11

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