class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:

        #The straightforward stragtegy is to create a Set to store what appears before.
        #However, ti takes O(n) space
        #The trick used here is to the nums itself as a Set

        #Time complexity: O(n)
        #Space complexity: O(1)
        res = []

        for ele in nums:
            idx = abs(ele) - 1
            if nums[idx] < 0:
                res.append(abs(ele))
            else:
                nums[idx] *= -1

        return res