First Missing Positive

Given an unsorted integer array nums, return the smallest positive integer that does not appear in the array.
Your algorithm must run in O(n) time and use O(1) extra space, excluding the input array.

Function
firstMissingPositive(nums: int[]) → int

Examples
Example 1
nums = [1,2,0]
return = 3
The positive values 1 and 2 are present, so 3 is the first missing positive.
Example 2
nums = [3,4,-1,1]
return = 2
The value 1 is present but 2 is absent.
Example 3
nums = [7,8,9,11,12]
return = 1
No value 1 appears.

Constraints
1 <= nums.length <= 100000.
-2^31 <= nums[i] <= 2^31 - 1.

The answer is always in the range 1 to n+1, where n is the array length. So use the array as its own hash table. Walk through it and, for each value v between 1 and n, swap it into index v-1, repeating until the current slot holds an out-of-range value or the right value already sits at its target. Then scan once: the first index i where nums[i] != i+1 gives the answer i+1. If all match, return n+1. The classic pitfall is an infinite loop on duplicates, like [1,1]. Always compare nums[nums[i]-1] to nums[i] before swapping, not just the index. Another trap is ignoring negatives and huge values near 2^31, which just get skipped. Sorting is O(n log n) and a set breaks the space rule, so both fail the stated constraints. If the swap loop slips under pressure, StealthCoder on the live OA is your hedge.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.