Count all pairs of an array which differ in K bits

Given an array of size n and integer k, count all pairs in array which differ in exactly K bits of binary representation of both the numbers.

The input arrays have elements with small values and possibly many repetitions.
Examples:

Input: arr[] = {2, 4, 1, 3, 1}
       k = 2       
Output: 5
Explanation:
There are only 4 pairs which differs in 
exactly 2 bits of binary representation:
(2, 4), (1, 2) [Two times] and (4, 1)
[Two times]

Input  : arr[] = {2, 1, 2, 1}
         k = 2
Output :  4

We strongly recommend that you click here and practice it, before moving on to the solution.

Naive Approach

A brute force is to run the two loops one inside the another and select the pairs one by one and take a XOR of both elements. The result of XORed value contains a set bits which are differ in both the elements. Now we just need to count total set bits so that we compare it with value K.

Below is the implementation of above approach:

C++

// C++ program to count all pairs with bit difference 
// as k 
#include<bits/stdc++.h> 
using namespace std; 
  
// Utility function to count total ones in a number 
int bitCount(int n) 
{ 
    int count = 0; 
    while (n) 
    { 
        if (n & 1) 
            ++count; 
        n >>= 1; 
    } 
    return count; 
} 
  
// Function to count pairs of K different bits 
long long countPairsWithKDiff(int arr[], int n, int k) 
{ 
    long long ans = 0; // initialize final answer 
  
    for (int i = 0; i < n-1; ++i) 
    { 
        for (int j = i + 1; j < n; ++j) 
        { 
            int xoredNum = arr[i] ^ arr[j]; 
  
            // Check for K differ bit 
            if (k == bitCount(xoredNum)) 
                ++ans; 
        } 
    } 
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    int k = 2; 
    int arr[] = {2, 4, 1, 3, 1}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
  
    cout << "Total pairs for k = " << k << " are "
         << countPairsWithKDiff(arr, n, k) << "\n"; 
  
    return 0; 
} 

Java

// Java program to count all pairs with bit difference 
// as k 
  
import java.io.*; 
  
class GFG { 
  
  
// Utility function to count total ones in a number 
static int bitCount(int n) 
{ 
    int count = 0; 
    while (n>0) 
    { 
        if ((n & 1)>0) 
            ++count; 
        n >>= 1; 
    } 
    return count; 
} 
  
// Function to count pairs of K different bits 
static long countPairsWithKDiff(int arr[], int n, int k) 
{ 
    long  ans = 0; // initialize final answer 
  
    for (int i = 0; i < n-1; ++i) 
    { 
        for (int j = i + 1; j < n; ++j) 
        { 
            int xoredNum = arr[i] ^ arr[j]; 
  
            // Check for K differ bit 
            if (k == bitCount(xoredNum)) 
                ++ans; 
        } 
    } 
    return ans; 
} 
  
// Driver code 
  
    public static void main (String[] args) { 
            int k = 2; 
    int arr[] = {2, 4, 1, 3, 1}; 
    int n =arr.length; 
  
    System.out.println( "Total pairs for k = " + k + " are "
        + countPairsWithKDiff(arr, n, k) + "\n"); 
    } 
} 
// This code is contributed by shs.. 

Python3

# Python 3 program to count all pairs  
# with bit difference as k 
  
# Utility function to count total  
# ones in a number 
def bitCount(n): 
    count = 0
    while (n): 
        if (n & 1): 
            count += 1
        n >>= 1
  
    return count 
  
# Function to count pairs of K different bits 
def countPairsWithKDiff(arr, n, k): 
    ans = 0
      
    # initialize final answer 
    for i in range(0, n - 1, 1): 
        for j in range(i + 1, n, 1): 
            xoredNum = arr[i] ^ arr[j] 
  
            # Check for K differ bit 
            if (k == bitCount(xoredNum)): 
                ans += 1
  
    return ans 
  
# Driver code 
if __name__ == '__main__': 
    k = 2
    arr = [2, 4, 1, 3, 1] 
    n = len(arr) 
  
    print("Total pairs for k =", k, "are",  
           countPairsWithKDiff(arr, n, k)) 
  
# This code is contributed by 
# Sanjit_Prasad 

C#

// C# program to count all pairs  
// with bit difference as k 
using System; 
  
class GFG  
{ 
  
// Utility function to count  
// total ones in a number 
static int bitCount(int n) 
{ 
    int count = 0; 
    while (n > 0) 
    { 
        if ((n & 1) > 0) 
            ++count; 
        n >>= 1; 
    } 
    return count; 
} 
  
// Function to count pairs of K different bits 
static long countPairsWithKDiff(int []arr,  
                                int n, int k) 
{ 
    long ans = 0; // initialize final answer 
  
    for (int i = 0; i < n-1; ++i) 
    { 
        for (int j = i + 1; j < n; ++j) 
        { 
            int xoredNum = arr[i] ^ arr[j]; 
  
            // Check for K differ bit 
            if (k == bitCount(xoredNum)) 
                ++ans; 
        } 
    } 
    return ans; 
} 
  
// Driver code 
public static void Main ()  
{ 
    int k = 2; 
    int []arr = {2, 4, 1, 3, 1}; 
    int n = arr.Length; 
      
    Console.WriteLine( "Total pairs for k = " +  
                                  k + " are " +  
        countPairsWithKDiff(arr, n, k) + "\n"); 
} 
} 
  
// This code is contributed by shs.. 

PHP

<?php 
// PHP program to count all  
// pairs with bit difference 
// as k 
  
// Utility function to count 
// total ones in a number 
function bitCount($n) 
{ 
    $count = 0; 
    while ($n) 
    { 
        if ($n & 1) 
            ++$count; 
        $n >>= 1; 
    } 
    return $count; 
} 
  
// Function to count pairs 
// of K different bits 
function countPairsWithKDiff($arr, $n, $k) 
{ 
      
    // initialize final answer 
    $ans = 0;  
  
    for ($i = 0; $i < $n-1; ++$i) 
    { 
        for ($j = $i + 1; $j < $n; ++$j) 
        { 
            $xoredNum = $arr[$i] ^ $arr[$j]; 
  
            // Check for K differ bit 
            if ($k == bitCount($xoredNum)) 
                ++$ans; 
        } 
    } 
    return $ans; 
} 
  
    // Driver code 
    $k = 2; 
    $arr = array(2, 4, 1, 3, 1); 
    $n = count($arr); 
  
    echo "Total pairs for k = " , $k , " are "
        , countPairsWithKDiff($arr, $n, $k) , "\n"; 
  
// This code is contributed by anuj_67. 
?> 

Output:

Total pairs for k = 2 are 5

Time complexity: O(N² * log MAX) where MAX is maximum element in input array.
Auxiliary space: O(1)

Efficient approach

This approach is efficient for the cases when input array has small elements and possibly many repetitions. The idea is to iterate from 0 to max(arr[i]) and for every pair(i, j) check the number of set bits in (i ^ j) and compare this with K. We can use inbuilt function of gcc( __builtin_popcount ) or precompute such array to make the check faster. If number of ones in i ^ j is equals to K then we will add the total count of both i and j.

C++

// Below is C++ approach of finding total k bit 
// difference pairs 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to calculate K bit different pairs in array 
long long kBitDifferencePairs(int arr[], int n, int k) 
{ 
    // Get the maximum value among all array elemensts 
    int MAX = *max_element(arr, arr+n); 
  
    // Set the count array to 0, count[] stores the 
    // total frequency of array elements 
    long long count[MAX+1]; 
    memset(count, 0, sizeof(count)); 
  
    for (int i=0; i < n; ++i) 
        ++count[arr[i]]; 
  
    // Initialize result 
    long long ans = 0; 
  
    // For 0 bit answer will be total count of same number 
    if (k == 0) 
    { 
        for (int i = 0; i <= MAX; ++i) 
            ans += (count[i] * (count[i] - 1)) / 2; 
  
        return ans; 
    } 
  
  
    for (int i = 0; i <= MAX; ++i) 
    { 
        // if count[i] is 0, skip the next loop as it 
        // will not contribute the answer 
        if (!count[i]) 
           continue; 
  
        for (int j = i + 1; j <= MAX; ++j) 
        { 
            //Update answer if k differ bit found 
            if ( __builtin_popcount(i ^ j) == k) 
                ans += count[i] * count[j]; 
        } 
    } 
    return ans; 
} 
  
// Driver code 
int main() 
{ 
    int k = 2; 
    int arr[] = {2, 4, 1, 3, 1}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
  
    cout << "Total pairs for k = " << k << " are = "
         << kBitDifferencePairs(arr, n, k) << "\n"; 
  
    k = 3; 
    cout << "Total pairs for k = " << k << " are = "
         << kBitDifferencePairs(arr, n, k) ; 
    return 0; 
} 

Java

// Below is Java approach of finding total k bit  
// difference pairs  
import java.util.*; 
  
class GFG  
{ 
  
// Function to calculate K bit  
// different pairs in array  
static long kBitDifferencePairs(int arr[], 
                                int n, int k)  
{  
    // Get the maximum value among all array elemensts  
    int MAX = Arrays.stream(arr).max().getAsInt();  
  
    // Set the count array to 0,  
    // count[] stores the total  
    // frequency of array elements  
    long []count = new long[MAX + 1];  
    Arrays.fill(count, 0); 
  
    for (int i = 0; i < n; ++i)  
        ++count[arr[i]];  
  
    // Initialize result  
    long ans = 0;  
  
    // For 0 bit answer will be total 
    // count of same number  
    if (k == 0)  
    {  
        for (int i = 0; i <= MAX; ++i)  
            ans += (count[i] * (count[i] - 1)) / 2;  
  
        return ans;  
    }  
  
    for (int i = 0; i <= MAX; ++i)  
    {  
        // if count[i] is 0, skip the next loop  
        // as it will not contribute the answer  
        if (count[i] == 0)  
        continue;  
  
        for (int j = i + 1; j <= MAX; ++j)  
        {  
            // Update answer if k differ bit found  
            if ( Integer.bitCount(i ^ j) == k)  
                ans += count[i] * count[j];  
        }  
    }  
    return ans;  
}  
  
// Driver code  
public static void main(String[] args) 
{ 
    int k = 2;  
    int arr[] = {2, 4, 1, 3, 1};  
    int n = arr.length;  
  
    System.out.println("Total pairs for k = " +  
                                k + " are = " +  
                        kBitDifferencePairs(arr, n, k));  
  
    k = 3;  
    System.out.println("Total pairs for k = " + 
                                k + " are = " +  
                        kBitDifferencePairs(arr, n, k));  
    } 
}  
  
// This code is contributed by Rajput-Ji 

Python3

# Below is Python3 approach of finding  
# total k bit difference pairs 
  
# Function to calculate K bit different 
# pairs in array 
def kBitDifferencePairs(arr, n, k): 
  
    # Get the maximum value among  
    # all array elemensts 
    MAX = max(arr) 
  
    # Set the count array to 0, count[] stores  
    # the total frequency of array elements 
    count = [0 for i in range(MAX + 1)] 
  
    for i in range(n): 
        count[arr[i]] += 1
  
    # Initialize result 
    ans = 0
  
    # For 0 bit answer will be total  
    # count of same number 
    if (k == 0): 
        for i in range(MAX + 1): 
            ans += (count[i] * (count[i] - 1)) // 2
  
        return ans 
  
  
    for i in range(MAX + 1): 
          
        # if count[i] is 0, skip the next loop  
        # as it will not contribute the answer 
        if (count[i] == 0): 
            continue
  
        for j in range(i + 1, MAX + 1): 
              
            # Update answer if k differ bit found 
            if (bin(i ^ j).count('1') == k): 
                ans += count[i] * count[j] 
      
    return ans 
  
# Driver code 
k = 2
arr = [2, 4, 1, 3, 1] 
n = len(arr) 
  
print("Total pairs for k =", k, "are",  
       kBitDifferencePairs(arr, n, k)) 
  
k = 3
print("Total pairs for k =", k, "are",  
       kBitDifferencePairs(arr, n, k))  
  
# This code is contributed by mohit kumar 

C#

// Below is C# approach of finding  
// total k bit difference pairs  
using System; 
using System.Linq; 
  
class GFG  
{ 
  
// Function to calculate K bit  
// different pairs in array  
static long kBitDifferencePairs(int []arr, 
                                int n, int k)  
{  
    // Get the maximum value among  
    // all array elemensts  
    int MAX = arr.Max();  
  
    // Set the count array to 0,  
    // count[] stores the total  
    // frequency of array elements  
    long []count = new long[MAX + 1];  
  
    for (int i = 0; i < n; ++i)  
        ++count[arr[i]];  
  
    // Initialize result  
    long ans = 0;  
  
    // For 0 bit answer will be total 
    // count of same number  
    if (k == 0)  
    {  
        for (int i = 0; i <= MAX; ++i)  
            ans += (count[i] *   
                   (count[i] - 1)) / 2;  
  
        return ans;  
    }  
  
    for (int i = 0; i <= MAX; ++i)  
    {  
        // if count[i] is 0, skip the next loop  
        // as it will not contribute the answer  
        if (count[i] == 0)  
        continue;  
  
        for (int j = i + 1; j <= MAX; ++j)  
        {  
            // Update answer if k differ bit found  
            if (BitCount(i ^ j) == k)  
                ans += count[i] * count[j];  
        }  
    }  
    return ans;  
}  
  
static int BitCount(int n)  
{  
    int count = 0;  
    while (n > 0) 
    {  
        count += n & 1;  
        n >>= 1;  
    }  
    return count;  
}  
  
// Driver code  
public static void Main(String[] args) 
{ 
    int k = 2;  
    int []arr = {2, 4, 1, 3, 1};  
    int n = arr.Length;  
  
    Console.WriteLine("Total pairs for k = " +  
                               k + " are = " +  
                       kBitDifferencePairs(arr, n, k));  
  
    k = 3;  
    Console.WriteLine("Total pairs for k = " + 
                               k + " are = " +  
                       kBitDifferencePairs(arr, n, k));  
    } 
}  
  
// This code is contributed by PrinciRaj1992 

Output:

Total pairs for k = 2 are = 5

Time complexity: O(MAX²) where MAX is maximum element in input array.
Auxiliary space: O(MAX)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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