That's a great work around.
Would you be able to do it in O(n) time? numbers.sort usually be O(n log n) (unless using TimSort and data is already sorted).
O(n)
numbers.sort
O(n log n)
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That's a great work around.
Would you be able to do it in
O(n)time?numbers.sortusually beO(n log n)(unless using TimSort and data is already sorted).