41. First Missing Positive

Given an unsorted integer array nums. Return the smallest positive integer that is not present in nums.

You must implement an algorithm that runs in O(n) time and uses O(1) auxiliary space.

Example 1:

Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.

Example 2:

Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1

Approach 01:

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        unordered_map<int,int> mp;
        for(auto &it: nums){
            mp[it] = 1;
        }

        int n = mp.size();
        for(int i=1; i<n+2; i++){
            if(mp[i] != 1){
                return i;
            }
        }
        return -1;
    }
};

from typing import *
import collections

class Solution:
    def firstMissingPositive(self, nums: List[int]) -> int:
        d = collections.Counter(nums)
        i = 0

        for i in range(1,len(d)+2):
            if(d.get(i) == None):
                return i
        
        return -1

Time Complexity:

Building the Counter: This operation takes O(n) time, where n is the number of elements in the nums list.

For Loop: The range of the loop runs from 1 to len(d) + 1. In the worst case, len(d) = n, so the loop will iterate up to n + 1 times. Each iteration performs a dictionary get operation, which is O(1) on average.

Combining these, the total time complexity is O(n).

Space Complexity:

Space for the Counter: In the worst case, the Counter will store each element as a key, requiring O(n) space.

Additional variables take O(1) space.

Combining these, the total space complexity is O(n).