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In the code snippet (which will not pass due to time constraints) word does not exist in T until it's loop checking against the current set of T is complete and it is added at the end. See the line after the T loop to add the current word to T.
It is not possible to have T and word refer to the same element at the same time. nless you mean there is a duplicate word (and therefore the set fails).
You could think of word as a "yet to check" list and T as a "has been checked " list
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In your solution, how to avoid the situation when 'work' and 't' are exactly the same elements?
In the code snippet (which will not pass due to time constraints) word does not exist in T until it's loop checking against the current set of T is complete and it is added at the end. See the line after the T loop to add the current word to T.
It is not possible to have T and word refer to the same element at the same time. nless you mean there is a duplicate word (and therefore the set fails).
You could think of word as a "yet to check" list and T as a "has been checked " list