For each possible pair `(i, j)`, compute the bitwise AND `nums[i] & nums[j]` and store the frequency of each result in a hash map. Then, for each `nums[k]`, check if `nums[k]` can form a triple with any of these AND results to make the final AND zero (i.e., `(nums[i] & nums[j]) & nums[k] == 0`).

Optimize the previous approach by precomputing all possible AND results of pairs and their frequencies. Then, for each `nums[k]`, iterate through the hash map and count how many pairs `(i, j)` satisfy `(pair_and & nums[k]) == 0`. Multiply this count by the frequency of `nums[k]` to account for all possible `k` values.

Further optimize by leveraging the fact that the AND operation is monotonic (i.e., `a & b <= a` and `a & b <= b`). Precompute all possible AND results of pairs and their frequencies, then use inclusion-exclusion or bitmask techniques to efficiently count the number of valid triples without iterating through all possible `k` values explicitly. This reduces the time complexity from O(n³) to O(n² + m * 2^b), where `m` is the number of unique AND results and `b` is the number of bits in the integers.