Is Find All Duplicates in an Array easy or hard?

The problem is generally classified as Medium difficulty. The challenge comes from recognizing that the value range 1..n allows in-place marking or index placement, enabling an O(n) time and O(1) space solution instead of using extra memory.

Find All Duplicates in an Array Python/Java solution

In Python or Java, the typical implementation iterates through the array and uses negation marking. For each number x, compute index x-1 and flip the sign of nums[index]. If nums[index] is already negative, x is a duplicate and should be added to the result list.

How to solve Find All Duplicates in an Array in O(n)?

Use the negation marking technique. Iterate through the array, compute the index as abs(nums[i]) - 1, and negate the value at that index if it is positive. If the value at that index is already negative, the number has appeared before and should be added to the result list.

What is the best approach for Find All Duplicates in an Array?

Negation marking is the most efficient and commonly expected approach. It runs in O(n) time and O(1) extra space by using the array itself as a visitation map. Each number marks its corresponding index by flipping the sign, and encountering an already negative value indicates a duplicate.

Is Find All Duplicates in an Array asked at Google/Amazon/Meta?

Array manipulation and in-place marking problems like this appear frequently in interviews at companies such as Amazon, Meta, and Google. The question tests understanding of array indexing tricks, space optimization, and recognizing constraints like values limited to 1..n.

What data structure is used in Find All Duplicates in an Array?

The optimized solutions primarily rely on the array itself as the tracking structure. Instead of allocating a hash set or hash map, the algorithm marks visited indices by modifying the array values. This reduces auxiliary memory usage while preserving linear time complexity.

What is the time complexity of Find All Duplicates in an Array?

The optimal solutions run in O(n) time because each element is processed at most once or twice. Both negation marking and index-based value reordering achieve linear traversal of the array. Space complexity can be reduced to O(1) by modifying the array in-place.

442. Find All Duplicates in an Array - Practice Online

Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears at most twice, return an array of all the integers that appears twice.

You must write an algorithm that runs in O(n) time and uses only constant auxiliary space, excluding the space needed to store the output

Example 1:

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

Example 2:

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

Example 3:

Input: nums = [1]
Output: []

Constraints:

n == nums.length
1 <= n <= 10⁵
1 <= nums[i] <= n
Each element in nums appears once or twice.