First missing positive

Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses constant extra space.

Example 1:

Input: nums = [1,2,0]
Output: 3

Example 2:

Input: nums = [3,4,-1,1]
Output: 2

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1

Constraints:

• 1 <= nums.length <= 5 * 105
• -2^31 <= nums[i] <= 2^31 - 1

Approach 1:

Run a loop from 1 to len(nums)+1 including both . Then for each time , check if the number is present in the nums array or not .

loop from 1 to length of nums+1:
      check if the number is present in the nums array
      if number  not present in nums :
          return number

Time: O(n²)

Space:O(1)

Approach 2:

Using hash-map

Using hash map with the above technique considerably reduces the cost of traversing the array each time

class Solution(object):
    def firstMissingPositive(self, nums):
        val=set(nums)
        for i in range(1,len(nums)+2):
            
            if i not in val and i>0:
                return i

Time: O(n)

Space:O(n)

Approach 3 :

Now to a much more optimized solution.

Here we assume that the number is to be positioned to an index , number - 1 . Iterate through the array and for each of the number currently being explored , swap it with the number at its actual index position. While traversing, if a number is at its correct index position just leave that number and move to the next index position.

class Solution(object):
    def firstMissingPositive(self, nums):
        i=0
        while(i<len(nums)):
            if nums[i]<=0 or nums[i]>len(nums):
                i+=1
            else:
                if nums[i]==i+1:
                    i+=1
                else:
                    if nums[nums[i]-1]==nums[i]:
                        i+=1
                    else:
                        temp=nums[nums[i]-1]    
                        nums[nums[i]-1]=nums[i]
                        nums[i]=temp
        j=0
        while(j<len(nums)):
            if(j+1!=nums[j]):
                return(j+1)
            j+=1
        return(len(nums)+1)

Time: O(n)

Space: O(1)