Single Number II

Problem: Every element appears three times except one. Find it.

Approach: Count bits. For each bit position, sum the bits across all numbers. If divisible by $3$, the unique element has $0$ there. Otherwise, it has $1$.

function singleNumber(nums):
    result = 0
    for i from 0 to 31:
        bitSum = 0
        for num in nums:
            bitSum = bitSum + ((num >> i) & 1)
        if bitSum mod 3 != 0:
            result = result | (1 << i)
    return result

Time: $O(32n) = O(n)$.

Space: $O(1)$ (excluding input array).

Alternative: Use two variables to simulate a base-$3$ counter per bit.