Count all pairs of an array which differ in K bits
Last Updated :
13 Sep, 2023
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 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..
# 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# 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 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.
?>
<script>
// JavaScript program to count all pairs with bit difference
// as k
// Utility function to count total ones in a number
function bitCount(n)
{
let count = 0;
while (n>0)
{
if ((n & 1)>0)
++count;
n >>= 1;
}
return count;
}
// Function to count pairs of K different bits
function countPairsWithKDiff(arr, n, k)
{
let ans = 0; // initialize final answer
for (let i = 0; i < n-1; ++i)
{
for (let j = i + 1; j < n; ++j)
{
let xoredNum = arr[i] ^ arr[j];
// Check for K differ bit
if (k == bitCount(xoredNum))
++ans;
}
}
return ans;
}
// Driver Code
let k = 2;
let arr = [2, 4, 1, 3, 1];
let n = arr.length;
document.write( "Total pairs for k = " + k + " are "
+ countPairsWithKDiff(arr, n, k) + "\n");
</script>
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 elements
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;
}
// 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 elements
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
# 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 elements
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
// 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 elements
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
<script>
// Below is Javascript approach
// of finding total k bit
// difference pairs
// Function to calculate K bit
// different pairs in array
function kBitDifferencePairs(arr, n, k)
{
// Get the maximum value among
// all array elements
let MAX = Math.max(...arr);
// Set the count array to 0, count[] stores the
// total frequency of array elements
let count = new Array(MAX+1).fill(0);
for (let i=0; i < n; ++i)
++count[arr[i]];
// Initialize result
let ans = 0;
// For 0 bit answer will be total
// count of same number
if (k == 0)
{
for (let i = 0; i <= MAX; ++i)
ans += parseInt((count[i] *
(count[i] - 1)) / 2);
return ans;
}
for (let 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 (let 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;
}
function BitCount(n)
{
let count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Driver code
let k = 2;
let arr = [2, 4, 1, 3, 1];
let n = arr.length;
document.write("Total pairs for k = " + k + " are = "
+ kBitDifferencePairs(arr, n, k) + "<br>");
k = 3;
document.write("Total pairs for k = " + k + " are = "
+ kBitDifferencePairs(arr, n, k) + "<br>");
</script>
Output:
Total pairs for k = 2 are = 5
Time complexity: O(MAX2) where MAX is maximum element in input array.
Auxiliary space: O(MAX)
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