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# Good Pairs CodeChef Solution

## Problem – Good Pairs CodeChef Solution

You are given N integers: A1A2, …, AN. You need to count the number of pairs of indices (ij) such that 1 ≤ i < j ≤ N and Ai | Aj ≤ max(AiAj).

Note: Ai | Aj refers to bitwise OR.

### Input

• The first line of the input contains an integer T, denoting the number of test cases. The description of each testcase follows.
• The first line of each testcase contains a single integer: N
• The second line of each testcase contains N integers: A1A2, …, AN.

### Output

For each test case, output a single line containing the answer for that test case.

### Constraints

• 1 ≤ T ≤ 20
• 1 ≤ N ≤ 106
• 0 ≤ Ai ≤ 106
• 1 ≤ Sum of N over all test cases ≤ 106

• 1 ≤ N ≤ 103

• Original constraints

### Sample 1:

``````Input:
1
3
1 2 3
Output:
2``````

### Explanation:

There are three possible pairs of indices which satisfy 1 ? i < j ? N: (1, 2), (1, 3) and (2, 3). Let us see which of those satisfy Ai | Aj ? max(AiAj):

• (1, 2): A1 | A2 = 1 | 2 = (01)2 | (10)2 = (11)2 = 3. But max(A1A2) = max(1, 2) = 2, and 3 ? 2 is not correct. Hence this is not a valid pair.
• (1, 3): A1 | A3 = 1 | 3 = (01)2 | (11)2 = (11)2 = 3. And max(A1A3) = max(1, 3) = 3, and 3 ? 3 is correct. Hence this is a valid pair.
• (2, 3): A2 | A3 = 2 | 3 = (10)2 | (11)2 = (11)2 = 3. And max(A2A3) = max(2, 3) = 3, and 3 ? 3 is correct. Hence this is a valid pair.

So there are a total of 2 valid pairs, and hence the answer is 2.

## Good Pairs CodeChef Solution in Pyth 3

``````for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().strip().split()))
d = {}
for i in range(n):
if arr[i] not in d:
d[arr[i]] = 1
else:
d[arr[i]] += 1
ans = 0
for A1 in d.keys():
ans += d[A1] * (d[A1] - 1)
for A2 in d.keys():
if A1 != A2 and (A1 | A2) <= max(A1, A2):
ans += d[A1] * d[A2]
print(ans // 2)``````

## Good Pairs CodeChef Solution in C++14

``````#include<bits/stdc++.h>
#include<ostream>
using namespace std;

#define int long long
#define sz(x) ((int)x.size())
#define all(x) (x).begin(), (x).end()
const int INF = numeric_limits<int>::max();
const int nax = (int)(3000901);
const int mod = 1e9 + 7;

template<class X, class Y>
bool maximize(X& x, const Y y) {
if (y > x) {x = y; return true;}
return false;
}
template<class X, class Y>
bool minimize(X& x, const Y y) {
if (y < x) {x = y; return true;}
return false;
}

int arr[nax], dulpet[nax];
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
memset(dulpet, 0, sizeof dulpet);
for (int i = 1; i <= n; ++i) {
cin >> arr[i];
dulpet[arr[i]]++;
}
int ans = 0;
}
for (int i = 0; i < 20; ++i) {
if (mask & (1 << i)) {
}
}
}
for (int i = 1; i <= n; ++i) {
ans += dulpet[arr[i]] - 1;
}
cout << ans << '\n';
}
return 0;
}``````
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