The example they give looks simple enough, [3, 4, -1, 1] returns 2, sure fine.
Use the array itself as a hash table by placing each number x in its correct position (index x-1) through cyclic swaps. Then scan the array to find the first index i where nums[i] != i+1; the missing positive is i+1. If all are in place, the answer is n+1.
Pro tip: Clarify upfront that you're treating the array as a hash map and that the swap-based approach is O(n) because each element is moved at most once. Mention that you ignore non-positive numbers and values greater than n, as they can't be the answer.
Confirm the array can be modified, that extra space must be O(1), and discuss edge cases like empty array, all negatives, or all numbers present.
Describe how to place each positive integer x (where 1 <= x <= n) at index x-1 by swapping until the array is 'sorted' in terms of positions.
Iterate through the array; for each index i, while nums[i] is in range and not already in its correct position, swap nums[i] with nums[nums[i]-1].
After rearranging, scan the array from left to right; the first index i where nums[i] != i+1 gives the missing positive i+1. If none, return n+1.
Argue that each element is swapped at most once, so total swaps are O(n), and no extra space is used beyond a few variables.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.