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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
#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
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