Given an array A[] of n integers, find out the number of ordered pairs such that A_{i}&A_{j} is zero, where 0<=(i,j)<n. Consider (i, j) and (j, i) to be different.

**Constraints: **

1<=n<=10^{4}

1<=A_{i}<=10^{4}

Examples:

Input : A[] = {3, 4, 2} Output : 4 Explanation : The pairs are (3, 4) and (4, 2) which are counted as 2 as (4, 3) and (2, 4) are considered different. Input : A[]={5, 4, 1, 6} Output : 4 Explanation : (4, 1), (1, 4), (6, 1) and (1, 6) are the pairs

**Simple approach **: A simple approach is to check for all possible pairs and count the number of ordered pairs whose bitwise & returns 0.

Below is the implementation of above idea:

## C++

// CPP program to calculate the number // of ordered pairs such that thier bitwise // and is zero #include <bits/stdc++.h> using namespace std; // Naive function to count the number // of ordered pairs such that their // bitwise and is 0 int countPairs(int a[], int n) { int count = 0; // check for all possible pairs for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) if ((a[i] & a[j]) == 0) // add 2 as (i, j) and (j, i) are // considered different count += 2; } return count; } // Driver Code int main() { int a[] = { 3, 4, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << countPairs(a, n); return 0; }

## Java

// Java program to calculate the number // of ordered pairs such that thier bitwise // and is zero class GFG { // Naive function to count the number // of ordered pairs such that their // bitwise and is 0 static int countPairs(int a[], int n) { int count = 0; // check for all possible pairs for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) if ((a[i] & a[j]) == 0) // add 2 as (i, j) and (j, i) are // considered different count += 2; } return count; } // Driver Code public static void main(String arg[]) { int a[] = { 3, 4, 2 }; int n = a.length; System.out.print(countPairs(a, n)); } } // This code is contributed by Anant Agarwal.

## Python3

# Python3 program to calculate the number # of ordered pairs such that thier # bitwise and is zero # Naive function to count the number # of ordered pairs such that their # bitwise and is 0 def countPairs(a, n): count = 0 # check for all possible pairs for i in range(0, n): for j in range(i + 1, n): if (a[i] & a[j]) == 0: # add 2 as (i, j) and (j, i) are # considered different count += 2 return count # Driver Code a = [ 3, 4, 2 ] n = len(a) print (countPairs(a, n)) # This code is contributed # by Shreyanshi Arun.

## C#

// C# program to calculate the number // of ordered pairs such that thier // bitwise and is zero using System; class GFG { // Naive function to count the number // of ordered pairs such that their // bitwise and is 0 static int countPairs(int []a, int n) { int count = 0; // check for all possible pairs for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) if ((a[i] & a[j]) == 0) // add 2 as (i, j) and (j, i) // arev considered different count += 2; } return count; } // Driver Code public static void Main() { int []a = { 3, 4, 2 }; int n = a.Length; Console.Write(countPairs(a, n)); } } // This code is contributed by nitin mittal.

## PHP

<?php // PHP program to calculate the number // of ordered pairs such that their // bitwise and is zero // Naive function to count the number // of ordered pairs such that their // bitwise and is 0 function countPairs($a, $n) { $count = 0; // check for all possible pairs for ($i = 0; $i < $n; $i++) { for ($j = $i + 1; $j < $n; $j++) if (($a[$i] & $a[$j]) == 0) // add 2 as (i, j) and (j, i) are // considered different $count += 2; } return $count; } // Driver Code { $a = array(3, 4, 2); $n = sizeof($a) / sizeof($a[0]); echo countPairs($a, $n); return 0; } // This code is contributed by nitin mittal ?>

**Output:**

4

**Time Complexity:** O(n^{2})

**Efficient approach**: An efficient approach is to use Sum over Subsets Dynamic Programming method and count the number of ordered pairs. In the SOS DP we find out the pairs whose bitwise & returned 0. Here we need to count the number of pairs.

Some key observations are the constraints, the maximum that an array element can be is 10^{4}. Calculating the mask up to (1<<15) will give us our answer. Use hashing to count the occurrence of every element. If the last bit is OFF, then relating to SOS dp, we will have a base case since there is only one possibility of OFF bit.

dp[mask][0] = freq(mask)

If the last bit is set ON, then we will have the base case as:

dp[mask][0] = freq(mask) + freq(mask^1)

We add freq(mask^1) to add the other possibility of OFF bit.

**Iterate over N=15 bits, which is the maximum possible.**

Let’s consider the** i-th bit to be 0**, then no subset can differ from the mask in the i-th bit as it would mean that the numbers will have a 1 at i-th bit where the mask has a 0 which would mean that it is not a subset of the mask. Thus we conclude that the numbers now differ in the first (i-1) bits only. Hence,

DP(mask, i) = DP(mask, i-1)

Now the **second case, if the i-th bit is 1, ** it can be divided into two non-intersecting sets. One containing numbers with i-th bit as 1 and differing from mask in the next (i-1) bits. Second containing numbers with ith bit as 0 and differing from mask(2^{i}) in next (i-1) bits. Hence,

DP(mask, i) = DP(mask, i-1) + DP(mask2^{i}, i-1).

**DP[mask][i] **stores the number of subsets of mask which differ from mask only in first i bits. Iterate for all array elements, and for every array element add the number of subsets (dp[ ( ( 1<<N ) – 1 ) ^ a[i] ][ N ]) to the number of pairs. N = maximum number of bits.

**Explanation of addition of dp[ ( ( 1<<N ) – 1 ) ^ a[i] ][N] to the number of pairs: **Take an example of A[i] being 5, which is 101 in binary. For better understanding, assume N=3 in this case, therefore, the reverse of 101 will be 010 which on applying bitwise & gives 0. So (1<<3) gives 1000 which on subtraction from 1 gives 111. 111101 gives 010 which is the reversed bit.So dp[((1<<N)-1)^a[i]][N] will have the number of subsets that returns 0 on applying bitwise & operator.

Below is the implementation of the above idea:

// CPP program to calculate the number // of ordered pairs such that thier bitwise // and is zero #include <bits/stdc++.h> using namespace std; const int N = 15; // efficient function to count pairs long long countPairs(int a[], int n) { // stores the frequency of each number unordered_map<int, int> hash; long long dp[1 << N][N + 1]; memset(dp, 0, sizeof(dp)); // initialize 0 to all // count the frequency of every element for (int i = 0; i < n; ++i) hash[a[i]] += 1; // iterate for al possible values that a[i] can be for (long long mask = 0; mask < (1 << N); ++mask) { // if the last bit is ON if (mask & 1) dp[mask][0] = hash[mask] + hash[mask ^ 1]; else // is the last bit is OFF dp[mask][0] = hash[mask]; // iterate till n for (int i = 1; i <= N; ++i) { // if mask's ith bit is set if (mask & (1 << i)) { dp[mask][i] = dp[mask][i - 1] + dp[mask ^ (1 << i)][i - 1]; } else // if mask's ith bit is not set dp[mask][i] = dp[mask][i - 1]; } } long long ans = 0; // iterate for all the array element // and count the number of pairs for (int i = 0; i < n; i++) ans += dp[((1 << N) - 1) ^ a[i]][N]; // return answer return ans; } // Driver Code int main() { int a[] = { 5, 4, 1, 6 }; int n = sizeof(a) / sizeof(a[0]); cout << countPairs(a, n); return 0; }

Output:

4

**Time Complexity:** O(N*2^{N}) where N=15 which is maximum number of bits possible, since A_{max}=10^{4}.

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