Clarify the problem constraints (e.g., input size, integer range, handling of non-positive numbers) and then present an efficient bitwise solution using n > 0 && (n & (n - 1)) == 0. Walk through the logic, discuss time and space complexity, and mention edge cases like 0, 1, and negative numbers.
Pro tip: Mention that the bitwise trick works because powers of two have exactly one set bit, and subtracting 1 flips all bits below it, so the AND is zero. Also note that in languages like Java, you should use n > 0 to exclude negative numbers because of two's complement representation.
Ask about input size, integer range, whether the list can be empty, and how to handle 0 and negative numbers. Confirm the expected output format (e.g., list of 1s and 0s).
State that a positive integer is a power of two if and only if it has exactly one bit set. Show that n & (n - 1) clears the lowest set bit, so the result is 0 for powers of two.
Implement the function that iterates through the list and applies the condition n > 0 && (n & (n - 1)) == 0, returning 1 or 0 accordingly. Use clear variable names and handle edge cases.
State that time complexity is O(n) and space complexity is O(1) per element (or O(n) for the output). Walk through test cases: [1, 2, 3, 4, 0, -2] should yield [1,1,0,1,0,0].
Mention that a loop dividing by 2 repeatedly is O(log n) per number but less efficient; the bitwise method is optimal. Also note that for very large lists, parallelization or vectorization could be considered.
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