Triples with Bitwise AND Equal to Zero in C++

Suppose we have an array of integers A. We have to find the number of triples of indices (i, j, k) such that −

• 0 <= i < size of A

• 0 <= j < size of A

• 0 <= k < size of A

A[i] AND A[j] AND A[k] is 0, where AND represents the bitwise-AND operator.

So, if the input is like [3,1,2], then the output will be 12

• To solve this, we will follow these steps −

• Define one map m

• ret := 0

• n := size of A

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • for initialize j := 0, when j < n, update (increase j by 1), do −

    • for initialize j := 0, when j < n, update (increase j by 1), do −

• for initialize i := 0, when i < n, update (increase i by 1), do −

  • x := A[i]

  • for all key-value pairs a in m

    • if (a.key AND x) is same as 0, then −

      • ret := ret + a.value

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int countTriplets(vector<int>& A){
      unordered_map<int, int> m;
      int ret = 0;
      int n = A.size();
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < n; j++) {
            m[A[i] & A[j]]++;
         }
      }
      for (int i = 0; i < n; i++) {
         int x = A[i];
         for (auto& a : m) {
            if ((a.first & x) == 0) {
               ret += a.second;
            }
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {3,1,2};
   cout << (ob.countTriplets(v));
}

Input

{3,1,2}

Output

12