# Different Bits Sum Pairwise - InterviewBit Solution

**Problem**: __Different Bits Sum Pairwise__

### Problem Description:

We define f(X, Y) as a number of different corresponding bits in the binary representation of X and Y. For example, f(2, 7) = 2, since binary representation of 2 and 7 are 010 and 111, respectively. The first and the third bit differ, so f(2, 7) = 2.

You are given an array of N positive integers, A1, A2,…, AN. Find the sum of f(Ai, Aj) for all pairs (i, j) such that 1 ≤ i, j ≤ N. Return the answer modulo 109+7.

**For example,**

```
A=[1, 3, 5]
We return
f(1, 1) + f(1, 3) + f(1, 5) +
f(3, 1) + f(3, 3) + f(3, 5) +
f(5, 1) + f(5, 3) + f(5, 5) =
0 + 1 + 1 +
1 + 0 + 2 +
1 + 2 + 0 = 8
```

### Solution Approach:

### Solution in C++:

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int Solution::cntBits(vector<int> &A) { int ans=0; for(int i =0;i<A.size();i++) { for(int j=0;j<A.size();j++) { ans+=__builtin_popcount(A[i]^A[j]); ans=ans%1000000007; } } return ans; } why this is showing TLE in hight cases