Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

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

Solution Explanation for Find All Numbers Disappeared in an Array

This problem asks us to find all numbers within the range [1, n] that are missing from a given array nums of length n, where each element in nums is within the same range [1, n]. Two approaches are presented below, one using extra space and another using in-place manipulation.

Approach 1: Using a Set (or Boolean Array)

This approach leverages a set (or boolean array in languages like Java and C++) to track the presence of each number in the input array.

Algorithm:

1. Initialization: Create a set s (or a boolean array s of size n+1, initialized to false).
2. Mark Present Numbers: Iterate through nums. For each number x, mark its presence in s by adding it to the set (or setting s[x] to true).
3. Find Missing Numbers: Iterate from 1 to n. If a number i is not found in s (or s[i] is false), add it to the result list.
4. Return Result: Return the list of missing numbers.

Time Complexity: O(n) - We iterate through the array twice. Set operations (add and check) are typically O(1) on average. Space Complexity: O(n) - We use a set (or boolean array) of size n to store the present numbers.

Code Examples: (See the provided code snippets in Python, Java, C++, Go, and TypeScript)

Approach 2: In-place Manipulation (Without Extra Space)

This approach cleverly uses the input array itself to track the presence of numbers without requiring extra space. It modifies the array in-place.

Algorithm:

1. Use Array as a Hash Table: We use the array's indices as keys and the values to indicate presence.
2. Mark Presence: Iterate through the array. For each number x, find the index i = abs(x) - 1. If the number at that index nums[i] is positive, change its sign to negative. This indicates the presence of x. The use of abs(x) prevents overwriting the sign if a number has already been encountered.
3. Find Missing Numbers: After the first iteration, iterate through the array again. If nums[i] is positive, then the number i + 1 is missing. Add it to the result list.

Time Complexity: O(n) - We iterate through the array twice. Space Complexity: O(1) - We use constant extra space, not counting the output array.

Code Examples: (See the provided code snippets in Python, Java, C++, Go, and TypeScript). Note the use of abs() to handle potential negative numbers resulting from the in-place marking.

Choosing the Best Approach:

Approach 1 is simpler to understand and implement. Approach 2 is more efficient in terms of space complexity if space is a critical constraint, but it modifies the input array, which might not always be desirable. Choose the approach that best suits the specific requirements of your problem and coding style.