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Nice work, both methods solve the problem, but see my notes about the time and space complexities. Overall excellent work!

end
end

return list[0,new_array_length]

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This technically creates a new array, so it makes your space complexity O(n)

# Space Complexity: ?
# Time Complexity: O(n) - it will always run however long the list is.
# Space Complexity: O(1) - nothing new is created
def remove_duplicates(list)

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This could use a little drying up, but it works!

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2 participants