Given an array, nums, of $n$ integers where nums[i] is in the range $[1, n]$, return an array of all the integers in the range $[1, n]$ that doesn’t appear in nums.

Constraints:

• $n =$ nums.length

• $1 \leq n \leq 10^3$

• $1 \leq$ nums[i] $\leq n$

A cyclic sort is optimal for placing elements in their correct positions when dealing with an input array where elements range from $1$ to $n$. The strategy involves swapping elements iteratively until each number is positioned at its correct index. This method ensures that all elements are sorted in a single pass. The key idea is to rearrange the array so that each element $nums[i]$ is at the index $nums[i] - 1$.

After correctly positioning the elements, scan the array to detect any discrepancies. If an element does not match its expected index, it indicates a missing number.

The steps of the algorithm are given below:

Step 1: Place each element in its correct position

1. Iterate through the array from the first index (i = 0). For each element nums[i], determine its correct position in the array, which should be at index nums[i] - 1.

2. Check if the element is already in its correct position:

   1. If nums[i] is not equal to nums[nums[i] - 1], swap nums[i] with the element at its correct position. This ensures that the current element is moved to the correct index, and the element that was previously there is now at nums[i], ready to be checked and potentially moved.

   2. If nums[i] is already in the correct position (i.e., nums[i] == nums[nums[i] - 1]), simply move to the next index by incrementing i.

3. Continue this process until all elements are in their correct positions or until the entire array has been traversed.

Step 2: Identify missing numbers

1. Iterate through the array again.

2. For each index i, check if nums[i] equals i + 1.

   1. If nums[i] is not equal to i + 1, it means that i + 1 is missing from the array, so append i + 1 to the missing_numbers list.

Let’s look at the following illustration to get a better understanding of the solution: