Oh, I did it that way! Observe the following pattern (b_i represents the ith lower bit of n (1-based)):

Suppose that n has exactly one bit: then, your array will be b₁.

If n has two bits, the final array will be b₂, b₁, b₂.

If n has three bits, the final array will be b₃, b₂, b₃, b₁, b₃, b₂, b₃.

If n has four bits, the final array will be b₄, b₃, b₄, b₂, b₄, b₃, b₄, b₁, b₄, b₃, b₄, b₂, b₄, b₃, b₄.

You can see that in general, if the lowest significant bit of i is the k th one (1-based), then, the bit in the i th position (1-based) in the array of bits you get is b_{m + 1 - k}, where m is the number of bits n has. This can be formally proved with induction. (Hint: notice that the way the array is built is symmetrical).

Then, for this solution, you only need to iterate from l to r and find those bits. My submission: 24858226