Note that there are two cases for the count: if a_i + a_i is a power of 2, and if it isn't. In the first case, you have that you can pair a_i with the other cnt_ai - 1 elements for sum = 2a_i. (i.e. all of the indices that are not i, but their value equals a_i).If you do not substract 1 from cnt_ai, you would be pairing an element with itself. In the second case, you can pair it with cnt_sum - ai elements, as sum - a_i ≠ a_i. Then, you have no risk of pairing an element to itself. What the guy with the submission did was substracting 1 from cnt_ai before adding cnt_sum - ai to the answer in order that, if 2a_i = sum, ans only increases by the corresponding amount (cnt_ai - 1), and if it 2a_i ≠ sum, it increases by cnt_sum - ai. Therefore, both cases are handled at once. Finally, note that you are counting each valid pair twice. You are counting the pair (a, b) first when i = a, and then when i = b. In order for the algorithm to give the correct answer, you must account for this by dividing ans by 2.

I don't know why my this code gives TLE. I just used other way of counting by keeping the records of elements already looked. Could you help me please.