Count triples with Bitwise AND equal to Zero

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Given an array of integers A[] consisting of N integers, find the number of triples of indices (i, j, k) such that A[i] & A[j] & A[k] is 0(<0 ? i, j, k ? N and & denotes Bitwise AND operator.

Examples:

Input: A[]={2, 1, 3}
Output: 12
Explanation: The following i, j, k triples can be chosen whose bitwise AND is zero:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2

Input: A[]={21, 15, 6}
Output: 0
Explanation: No such triplets exist.

Approach: The idea to solve this problem is to use a Map to process the array solving elements. Follow the steps below to solve the problem:

Initialize a Map to store frequencies of every possible value of A[i] & A[j]. Also, initialize a variable answer with 0, to store the required count.
Traverse the array and for each array element, traverse the map and check for each map if key, if it’s Bitwise AND with the current array element is 0 or not. For every array element for which it is found to be true, increase answer by frequency of the key.
After completing the traversal of the array, print answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
int countTriplets(vector<int>& A)
{
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    unordered_map<int, int> tuples;
 
    // Traverse the array
    for (auto a : A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        for (auto b : A)
            ++tuples[a & b];
 
    // Traverse the array
    for (auto a : A)
 
        // Iterate the map
        for (auto t : tuples)
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.first & a) == 0)
                cnt += t.second;
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
int main()
{
 
    // Input Array
    vector<int> A = { 2, 1, 3 };
 
    // Function Call
    cout << countTriplets(A);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
static int countTriplets(int []A)
{
   
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    HashMap<Integer,Integer> tuples = new HashMap<Integer,Integer>();
 
    // Traverse the array
    for (int a : A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        for (int b : A) 
        {
            if(tuples.containsKey(a & b))
                tuples.put(a & b, tuples.get(a & b) + 1);
            else
                tuples.put(a & b, 1);
        }
 
    // Traverse the array
    for (int a : A)
 
        // Iterate the map
        for (Map.Entry<Integer, Integer> t : tuples.entrySet())
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.getKey() & a) == 0)
                cnt += t.getValue();
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Input Array
    int []A = { 2, 1, 3 };
 
    // Function Call
    System.out.print(countTriplets(A));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to find the number of
# triplets whose Bitwise AND is 0.
def countTriplets(A) :
 
    # Stores the count of triplets
    # having bitwise AND equal to 0
    cnt = 0;
 
    # Stores frequencies of all possible A[i] & A[j]
    tuples = {};
 
    # Traverse the array
    for a in A:
 
        # Update frequency of Bitwise AND
        # of all array elements with a
        for b in A:
            if (a & b) in tuples: 
                tuples[a & b] += 1;                
            else:
                tuples[a & b] = 1;
 
    # Traverse the array
    for a in A:
         
        # Iterate the map
        for t in tuples: 
 
            # If bitwise AND of triplet
            # is zero, increment cnt
            if ((t & a) == 0):
                cnt += tuples[t];
 
    # Return the number of triplets
    # whose Bitwise AND is 0.
    return cnt;
 
# Driver Code
if __name__ ==  "__main__" :
 
    # Input Array
    A = [ 2, 1, 3 ];
 
    # Function Call
    print(countTriplets(A));
 
    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
static int countTriplets(int []A)
{
   
    // Stores the count of triplets
    // having bitwise AND equal to 0
    int cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    Dictionary<int,int> tuples = new Dictionary<int,int>();
 
    // Traverse the array
    foreach (int a in A)
 
        // Update frequency of Bitwise AND
        // of all array elements with a
        foreach (int b in A) 
        {
            if(tuples.ContainsKey(a & b))
                tuples[a & b] = tuples[a & b] + 1;
            else
                tuples.Add(a & b, 1);
        }
 
    // Traverse the array
    foreach (int a in A)
 
        // Iterate the map
        foreach (KeyValuePair<int, int> t in tuples)
 
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((t.Key & a) == 0)
                cnt += t.Value;
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Input Array
    int []A = { 2, 1, 3 };
 
    // Function Call
    Console.Write(countTriplets(A));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to find the number of
// triplets whose Bitwise AND is 0.
function countTriplets(A)
{
    // Stores the count of triplets
    // having bitwise AND equal to 0
    var cnt = 0;
 
    // Stores frequencies of all possible A[i] & A[j]
    var tuples = new Map();
 
    // Traverse the array
    A.forEach(a => {
         
        // Update frequency of Bitwise AND
        // of all array elements with a
        A.forEach(b => {
             
            if(tuples.has(a & b))
                tuples.set(a & b, tuples.get(a & b)+1)
            else
                tuples.set(a & b, 1)
        });
    });
 
    // Traverse the array
    A.forEach(a => {
         
        // Update frequency of Bitwise AND
        // of all array elements with a
        tuples.forEach((value, key) => {
             
            // If bitwise AND of triplet
            // is zero, increment cnt
            if ((key & a) == 0)
                cnt += value;
        });
    });
     
 
    // Return the number of triplets
    // whose Bitwise AND is 0.
    return cnt;
}
 
// Driver Code
// Input Array
var A = [2, 1, 3];
// Function Call
document.write( countTriplets(A));
 
</script>

Output:

Time Complexity: O(max(M*N, N²)) where M is the maximum element present in the given array
Auxiliary Space: O(M)

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