Given a non-empty array of integers `nums`, every element appears twice except for one. Find that single one in linear runtime and constant space.

Optimal Bit Manipulation Approach (XOR):
- Leverage the mathematical properties of the bitwise XOR (^) operator: x ^ x = 0 and x ^ 0 = x.

def single_number(nums: list[int]) -> int:
    result = 0
    for num in nums:
        result ^= num
    return result

Complexity & Key Logic:
- Time Complexity: O(N) single linear scan over the array.
- Space Complexity: O(1) constant auxiliary space.
- Algorithm Details:
- XOR is commutative and associative: the order of operations does not matter (a ^ b ^ a = (a ^ a) ^ b = 0 ^ b = b).
- Every number that appears twice cancels itself out to 0.
- The single unique element remains as the final value in result.