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Unformatted text preview: p e l i s t [ ] , i n t l e f t , i n t r i g h t )
/ / P o s t c o n d i t i o n: l i s t e l e m e n t s b e t w e e n i n d e x l e f t a n d
// index right have been sorted into increasing order
if( left < right )
int pivotPosition =
p artition( list, left, right );
Q u i c k s o r t( l i s t , l e f t , p i v o t P o s i t i o n - 1 ) ;
Q u i c k s o r t( l i s t , p i v o t P o s i t i o n + 1 , r i g h t ) ;
} So when can we say that f(n) is O(n log n)?
CPSC 260 Quicksort Page 5 CPSC 260 Quicksort Page 6 Analysis of Quicksort Algorithm
In the ideal case, the array of length N is partitioned into two
equal pieces of length N/2. Each of these pieces is then
partitioned into two pieces of length N/4 etc.
N/4 1 N/4 N/4 1 Similarly it can be argued that on each level, the partitioning
operation requires O(N) work. 1 CPSC 260 Quicksort 1 Page 7 It can also be shown that in the average case the time
complexity of the Quicksort algorithm is O( N log N ).
Unfortunately, the Quicksort algorithm presented here can
degrade in the case when the partitioning always results in
one partition having size 0.
This occurs, for
example, when the
pivot element is
chosen to be the
first element in the
array and the array
is already sorted. To partition an array of length N requires us to compare
each element in the array with the pivot element for a total of
N – 1 comparisons – an O(N) operation.
We then partition two arrays of length N/2. Each array of
length N/2 requires ( N/2 – 1 ) comparisons for a total of
2 * ( N/2 – 1 ) comparisons – again an O(N) operation. N/2
N/4 Each level in the diagram above represents a recursive step
in the Quicksort algorithm. Notice that on each call to
Quicksort, all the “work” is done by the partitioning function. In this ideal case where at each step the array is partitioned
into two pieces of equal length, the total number of levels in
the tree is O( log N ). Hence, in this case the time
complexity of the Quicksort algorithm...
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