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.

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

**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.

// 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; }

Output:Total pairs for k = 2 are = 5

**Time complexity: **O(N^{2} * 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.

// Below is C++ approach of finding total k bit // difference pairs #include<bits/stdc++.h> using namespace std; // Function to calclate 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; }

Output:Total pairs for k = 2 are = 5

**Time complexity: **O(MAX^{2}) where MAX is maximum element in input array.

**Auxiliary space: **O(MAX)

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