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 & 2Input: 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 Codeint main(){ // Input Array vector<int> A = { 2, 1, 3 }; // Function Call cout << countTriplets(A); return 0;} |
Java
// Java program for the above approachimport 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 Codepublic 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 Codeif __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 approachusing 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 Codepublic 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 Arrayvar A = [2, 1, 3];// Function Calldocument.write( countTriplets(A));</script> |
Output:
12
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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