how to use counting sort algorithm to sort array

Lecture 3
Sorting and Selection

Comparison Sort
ree.
decision t
a
as
viewed
be
can
sort
comparison
a
Therefore,
parts.
two
into
sequence
input
of
ns
permutatio
all
partitions
comparison
Each
sorts.
comparison
are
Quicksort
and
Heapsort
Sort,
Merge
Sort,
Insertion
}.
,...,
,
{
sequence
input
about
n
informatio
order
gain
to
elements
between
s
comparison
only
use
we
sort,
comparison
a
In
2
1 n
a
a
a

Decision Tree
?
2
1 a
a 
?
3
2 a
a 
?
3
2 a
a  ?
3
1 a
a 
?
3
1 a
a 
yes
yes
yes
yes yes
3
,
2
,
1
no
no
no
no no
2
,
3
,
1 2
,
1
,
3
3
,
1
,
2
1
,
3
,
2 1
,
2
,
3

Running Time
).
lg
(
)
(
Thus,
1
!
2
!
2
satisfies
)
(
depth
its
leaves,
!
has
ree
decision t
the
Since
ree.
decision t
the
of
height)
(or
depth
the
is
time
running
case)
(worst
the
sort,
comparison
a
In
1
12
)
(
n
n
n
T
e
n
e
n
e
n
e
e
n
n
n
n
T
n
n
n
n
n
n
T







 




















Is there a faster sorting of other type?

for
-
end
;
1
]]
[
[
]]
[
[
];
[
]]]
[
[
[
begin
do
1
downto
]
[
for
];
1
[
]
[
]
[
do
to
2
for
;
1
]]
[
[
]]
[
[
do
]
[
to
1
for
;
0
]
[
do
to
1
for













j
A
C
j
A
C
j
A
j
A
C
B
A
length
j
i
C
i
C
i
C
k
i
j
A
C
j
A
C
A
length
j
i
C
k
i
Counting sort
4
1,
4,
3,
1,
4,
6,
3,
:
A
1
0,
3,
2,
0,
2,
:
C
8
7,
7,
4,
2,
2,
:
C
4,
,
,
,
,
,
,
:
B
8
7,
6,
4,
2,
2,
:
C
4̂
1,
4,
3,
1,
4,
6,
3,
:
A

4
,
1̂
4,
3,
1,
4,
6,
3,
:
A
4,
,
,
,
,
1,
,
:
B
8
7,
6,
4,
2,
1,
:
C
4
1,
,
4̂
3,
1,
4,
6,
3,
:
A
4,
4,
,
,
,
1,
,
:
B
8
7,
5,
4,
2,
1,
:
C
4
1,
4,
,
3̂
1,
4,
6,
3,
:
A
4,
4,
,
3,
,
1,
,
:
B
8
7,
5,
3,
2,
1,
:
C
4
1,
4,
3,
,
1̂
4,
6,
3,
:
A
4,
4,
,
3,
,
1,
1,
:
B
8
7,
5,
3,
2,
0,
:
C

4
1,
4,
3,
1,
,
4̂
6,
3,
:
A
4,
4,
4,
3,
,
1,
1,
:
B
8
7,
4,
3,
2,
0,
:
C
4
1,
4,
3,
1,
4,
,
6̂
3,
:
A
6
4,
4,
4,
3,
,
1,
1,
:
B
7
7,
4,
3,
2,
0,
:
C
4
1,
4,
3,
1,
4,
6,
,
3̂
:
A
6
4,
4,
4,
3,
3,
1,
1,
:
B
7
7,
4,
2,
2,
0,
:
C

for
-
end
;
1
]]
[
[
]]
[
[
];
[
]]]
[
[
[
begin
do
1
downto
]
[
for
];
1
[
]
[
]
[
do
to
2
for
;
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]]
[
[
]]
[
[
do
]
[
to
1
for
;
0
]
[
do
to
1
for













j
A
C
j
A
C
j
A
j
A
C
B
A
length
j
i
C
i
C
i
C
k
i
j
A
C
j
A
C
A
length
j
i
C
k
i
Counting sort (Running Time)
)
(k
O
)
(n
O
)
(k
O
)
(n
O
)
( k
n
O 

for
-
end
;
1
do
0
]
[
while
;
1
]
[
]
[
;
]
[
begin
do
1
downto
]
[
for
];
[
;
1
]]
[
[
]]
[
[
do
]
[
to
1
for
;
0
]
[
do
to
1
for













k
k
k
C
k
C
k
C
k
j
B
A
length
j
C
length
k
j
A
C
j
A
C
A
length
j
i
C
k
i
A Student’s Variation
4
1,
4,
3,
1,
4,
6,
3,
:
A
1
0,
3,
2,
0,
2,
:
C
0
0,
3,
2,
0,
2,
:
C
6
,
,
,
,
,
,
,
:
B
6
4,
,
,
,
,
,
,
:
B
6
4,
4,
4,
,
,
,
,
:
B
6
4,
,
4
,
,
,
,
,
:
B
0
0,
0,
2,
0,
2,
:
C
6
4,
4,
4,
3,
,
,
,
:
B
6
4,
4,
4,
3,
3,
,
,
:
B

Counting sort is stable.
That is, the same value appear in the output
array in the same order as they do in the
input array.

Radix Sort
;
digit
on
array
sort
sort to
stable
a
use
"
do
to
1
for
)
,
(
Sort
-
Radix
digit.
of
number
a
is
element
each
,
array
input
In
i
A
d
i
d
A
d
A

355
720
436
839
657
457
329
839
329
657
457
436
355
720
657
457
355
839
436
329
720
839
720
657
457
436
355
329

Running Time
).
(
))
(
(
),
(
Since
.
to
1
range
in the
is
digit
each
if
))
(
(
dn
O
k
n
d
O
n
O
k
k
k
n
d
O





An Application:
Selection Problem
.
in
elements
1
exactly
n
larger tha
i.e.,
number,
smallest
th
the
is
which
,
:
Output
.
1
,
integer
an
and
numbers
distinct
of
set
A
:
Input
A
i
i
A
x
n
i
i
n
A





left.
the
th to
the
o
leftmost t
the
from
Count
2.
numbers.
input
Sort
.
1
i
)
( k
n
O 
)
(n
O
If we use comparison, what is the lower bound?
bound.
lower
a
is
)
(log
Therefore,
*)
,
...
,
,
,
(
...
)
...,
,
,
,*
(
)
...,
,
,
(*,
:
es
possibliti
only
are
There
n
O
n

Randomized Selection
);
1
,
,
2
,
(
Select
-
Randomized
return
else
]
1
[
return
then
,
1
if
else
)
,
,
,
(
Select
-
Randomized
return
then
if
;
1
);
,
,
(
Partition
-
Randomized
];
[
return
then
if
)
,
,
,
(
Select
Rondomized













k
i
r
q
A
q
A
k
i
i
q
p
A
k
i
p
q
k
r
p
A
q
p
A
r
p
i
r
p
A

Running time
 
   
)
4
(choose
.
4
3
)
1
2
/
)
1
)((
1
2
/
(
)
2
(
1
)]
(
E[
Then,
.
for
)]
(
E[
Assume
.)
)]
1
(
[
E
such that
(choose
.
)]
1
(
[
E
induction.
by
it
Prove
.
)]
(
[
E
Guess
.
0
some
for
)
))]
1
,
(max(
(E[
1
)]
(
E[
1
1
1
1
2
/
1
1
1
1
0
c
c
cn
n
c
n
c
n
c
n
n
n
n
n
c
n
c
ck
n
n
T
n
k
ck
k
T
c
T
c
c
T
cn
n
T
c
n
c
k
n
k
T
n
n
T
n
n
k
n
k
































Selection with O(n) Comparisons
 
 
.
if
side
high
on the
element
smallest
th
)
(
or the
,
if
side
low
on the
element
smallest
th
the
find
y to
recursivel
Select
Use
5.
partition.
the
of
side
low
on the
elements
of
number
the
be
Let
.
subroutine
Partition
apply
and
array
input
in
element
last
with the
Exchange
4.
medians.
group
n/5
of
median
the
find
y to
recursivel
Select
Use
3.
one.
larger
the
take
medians,
two
has
one
l
exceptiona
the
case,
In
sort.
insertion
by
group
each
of
median
the
Find
2.
elements.
5
than
less
of
one
except
possibly
elements,
5
of
groups
n/5
into
elements
Divide
.
1
:
Select
Algorithm
k
i
k
i
k
i
i
k
x
x
n




 
elements.
5
than
less
of
one
except
possibly
elements,
5
of
groups
n/5
into
elements
Divide
.
1 n
one.
smaller
the
take
medians,
two
has
one
l
exceptiona
the
case,
In
sort.
insertion
by
group
each
of
median
the
Find
2.
)
(n
O

 
medians.
group
n/5
of
median
the
find
y to
recursivel
Select
Use
3. x
 )
5
/
( n
T
x

partition.
the
of
side
low
on the
elements
of
number
the
be
Let
.
subroutine
Partition
apply
and
array
input
in
element
last
with the
Exchange
4.
k
x
x
)
,
1
,
(
Partition n
A
x
x
 
x
k
side
low side
high
)
(n
O

x
)
,
1
,
(
Partition n
A
x
x
 
x
k
side
low side
high
.
1
if
side
high
on the
element
smallest
th
)
1
(
or the
,
1
if
one
smallest
th
the
is
or
,
if
side
low
on the
element
smallest
th
the
find
y to
recursivel
Select
Use
5.
k
i
k
i
k
i
i
x
k
i
i







1

k
n
))
2
5
2
1
3
(
( 













n
n
T
Why? See next page!!!

( + low side) contains red part
( + high side) contains blue part















5
2
1
3
1
n
k
2
5
2
1
3 















n
k
n

Analysis
 
elements.
2
10
/
7
most
at
on
y
recursivel
called
is
Select
Thus,
deleted.
are
elements
2
10
3
2
5
2
1
3
least
at
iteration,
each
In
)
(
)
2
5
2
1
3
(
)
5
/
(
)
(
Therefore,









































n
n
n
n
O
n
n
T
n
T
n
T

Analysis
 
.
'
)
3
10
/
(
such that
enough
big
choose
n
whe
'
3
10
/
9
'
)
2
10
/
7
(
)
1
5
/
(
)
(
Then
.
for
)
(
Assume
.).
)
1
(
satisfy
to
(choose
)
1
(
induction.
by
it
prrove
Try
.
)
(
i.e.,
),
(
)
(
Guess
)
(
))
2
5
2
1
3
(
(
)
5
/
(
)
(
n
c
n
c
c
cn
n
c
c
cn
n
c
n
c
n
c
n
T
n
k
ck
k
T
c
T
c
c
T
cn
n
T
n
O
n
T
n
O
n
n
T
n
T
n
T


































basis
induction

).
59
/
)
59
(
,...,
2
/
)
2
(
),
1
(
,
'
20
max(
choose
ts,
requiremen
above
all
satisfy
To
.
59
)
59
(
,...,
2
)
2
(
,
)
1
(
have
should
we
step,
basis
at the
that
requires
This
'.
20
and
60
is,
That
.
'
)
20
/
(
and
20
/
3
10
/
that
so
,
3
20
/
have
to
sufficient
is
It
.
'
)
3
10
/
(
have
to
need
We
T
T
T
c
c
c
T
c
T
c
T
c
c
n
n
c
n
c
n
n
n
n
c
n
c













What we learnt in this lecture?
• What is “comparison sort”?.
• Lower bound for comparison sort.
• Counting sort and Radix sort.
• What is selection problem?
• expected linear time and linear time
selections.

Puzzle
median?
a
find
order to
in
done
be
to
has
s
comparison
many
how
case,
worst
in the
then
problem,
selection
solve
to
s
comparison
use
we
If

how to use counting sort algorithm to sort array

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