To solve this in O(1) space, we can use Bit Manipulation to count the number of 1-bits at each position (0-31). Since every number appears 3 times except for one, the sum of bits at each position modulo 3 will reveal the bits of the single number.

1. Initialize ans = 0.
2. Iterate through i from 0 to 31:
   • count = 0.
   • For every num in nums:
     • If the i-th bit of num is 1, count += 1.
   • count %= 3.
   • If count == 1, set the i-th bit of ans.
3. Handle negative numbers (Python integers are infinite, so bits are tricky).

• Time Complexity: O(32 * N) = O(N).
• Space Complexity: O(1).

def single_number(nums):
    ones, twos = 0, 0
    for n in nums:
        ones = (ones ^ n) & (~twos)
        twos = (twos ^ n) & (~ones)
    return ones

(Note: The ones/twos approach is a more elegant bitwise state-machine version of the bit-counting logic).