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| • |
ADT
ClearableStack
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– |
Standard
operations:
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push(),
pop(), isEmpty()
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Plus a
new operation:
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clear()
= remove everything
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from
the Stack
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Suppose
we implement
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clear()
by using multiple
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pops
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What
is the worst-case time
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for n
ClearableStack
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operations
starting with an
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empty
Stack?
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An
analysis:
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clear()
can take O(n) time in
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the
worst-case
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Thus,
worst-case time for n
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operations
is O(n2)
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A
better analysis
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There
is at most one pop()
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for
each insert()
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Thus
total time is
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proportional
to number of
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inserts
= O(n)
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Thus
the amortized time per
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operation
is O(1)
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