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The key insight is that AND being positive just means all elements in the subset share at least one common set bit.
The key insight is that the bitwise AND of a subset is greater than zero if and only if there exists at least one bit position that is set in every element of the subset. Therefore, for each bit position, count how many numbers have that bit set; the maximum count across all bit positions is the size of the largest valid subset. This reduces the problem to a single pass through the array while tracking bit counts.
Pro tip: Mention that this solution runs in O(n * B) time where B is the number of bits (e.g., 32 for integers), which is effectively O(n). Also, clarify that the subset can be any size, including singletons, and that the answer is at least 1 if the array is non-empty.
Realize that the bitwise AND of a subset is > 0 iff there is at least one bit position where all elements in the subset have that bit set. So we need to find the maximum number of elements sharing a common set bit.
Use an array of size equal to the number of bits (e.g., 32) to count how many numbers have each bit set. Alternatively, use a hash map if the bit range is unknown, but an array is simpler and faster.
For each number in the input array, iterate over its set bits (or all bit positions) and increment the corresponding counter. This can be done efficiently by checking each bit up to the maximum possible bit (e.g., 31 for 32-bit integers).
After processing all numbers, the answer is the maximum value in the bit-count array. If the array is empty, return 0; otherwise, the answer is at least 1.
Time complexity is O(n * B) where B is the number of bits (constant, e.g., 32), so effectively O(n). Space complexity is O(B) which is O(1).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.