An alternative solution if we consider that the sum of numbers: 1, 2, 3, ..., n is 1 + 2 + 3 + ... + n = n * (n+1) / 2, would be to sum all numbers in the loop and return n*(n+1)/2 - sum. Of course it has the same time complexity O(N) and and space complexity O(1) but it's only one pass. I know that the specific lesson follows this approach to demonstrate the pattern and to group similar problems under the same umbrella. I just thought of that solution when saw the problem, that's why I'm posting it here. Quick python code: {code} def find_missing_number(nums): n = len(nums) s = 0 for num in nums: s += num return (n * (n+1) // 2) - s {code}