algoadvance

1. Input Consistency: Is the input array always in the range 1 to n (inclusive), where n is the length of the array?
2. Duplicates: Can the input array contain duplicate numbers?
3. Output Requirements: Should the output be returned as a list of integers?

Strategy:

The problem requires finding all numbers in the range [1, n] that are missing from the given array. Given that the array might contain duplicates but the values are bound within 1 to n, we can use the properties of the array to mark which numbers are present.

Here’s a step-by-step strategy to achieve the solution efficiently:

1. Iterate through the array: Use the numbers in the array to mark corresponding indices (by making them negative) as visited.
2. Identify Missing Numbers: After marking, the indices which contain positive numbers indicate that their corresponding indices+1 values are missing in the array.
3. Result Compilation: Compile these missing values into the result list.

Code:

def findDisappearedNumbers(nums):
    # Mark numbers present in the array by marking the index corresponding to those numbers as negative
    for num in nums:
        index = abs(num) - 1
        if nums[index] > 0:
            nums[index] = -nums[index]

    # Collect the indices + 1 where the values are still positive
    result = []
    for i in range(len(nums)):
        if nums[i] > 0:
            result.append(i + 1)
    return result

Time Complexity:

• O(n): This is derived from the need to iterate through the array twice — first to mark the numbers and then to collect the missing numbers.
• Space Complexity: O(1), if we do not consider the output list. The algorithm modifies the input array in-place and uses only a constant amount of extra space.

This strategy effectively finds the missing numbers by leveraging the properties of the input array, ensuring an efficient and concise solution.

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