https://leetcode.com/problems/first-missing-positive

Notice, that first missinig positive will always be in range 1,2,...n,n+1, where n is length of nums. Let us rearrange numbers, putting each number on its place: number i on place i-1 (because indexes start with 0): let us iterate over our numbers and change two numbers if one of them not on its place: we break if number not in range 1,...,n or if we are trying to put number on the place, which is already occupied with this place (because we have infinite loop in this case)

When we iterate all numbers we find for number which is not on its place, using i == nums[i] - 1. It can happen that all numbers between 1 and n are here, so I add [0] to the end. Finally, I found index of False in this array: it will be our number.

Complexity: even though we have while loop inside for loop, complexity will be O(n): on each step we put at least one number to its proper place. Additional space complexity is O(1), however we modify our nums.

This solution is very similar to problem 442. Find All Duplicates in an Array

class Solution:
    def firstMissingPositive(self, nums):
        n = len(nums)
        for i in range(n):
            while nums[i]-1 in range(n) and nums[i] != nums[nums[i]-1]:
                nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]
                
        return next((i + 1 for i, num in enumerate(nums) if num != i + 1), n + 1)

If you like the solution, you can upvote it on leetcode discussion section: Problem 0041