Solution: Problem Challenge 3: Find the First K Missing Positive Numbers

class Solution:
  def findNumbers(self, nums, k):
    missing_numbers = []

    # swapping function
    def swap(i, j):
      nums[i], nums[j] = nums[j], nums[i]

    # if number not in position and number is not outside the array boundary and number does not point to number that is correctly positioned  
    for i in range(len(nums)):
      while nums[i] > 0 and nums[i] != i + 1 and nums[i] - 1 < len(nums) and nums[i] != nums[nums[i] - 1]:
        swap(i, nums[i] - 1)

    # find the missing numbers 
    wrong_numbers = set()
    for i in range(len(nums)):
      if nums[i] != i + 1:
        missing_numbers.append(i + 1)
        # add the miss-placed number to the set 
        wrong_numbers.add(nums[i])
        # if we have found all the missing k numbers just return
        

1. place all numbers at the correct index
2. store the wrongly indexed numbers in an object while reducing k
3. if k is greater than 0 add the remaining numbers to the missing array. to do this, you need to ensure its not in the store object which has all the numbers that already exist in the initial array that were wrongly indexed.

class Solution {
  findNumbers(nums, k) {
    let missingNumbers = [];
    let n = nums.length;
    let store = {}
    for (let i = 0; i < n; i++) {
      if (nums[i] > 0 && nums[i] <= n  && nums[i] !== nums[nums[i] - 1]) {
        [nums[nums[i] - 1], nums[i]] = [nums[i], nums[nums[i] - 1]];
      i--
      }
    }
    
    for (let i = 0; i < n; i++) {
      if (nums[i] !== i + 1 && k != 0){
          store[nums[i]] = 1 
          missingNumbe