Problem definition
Counting sort
Lecture 5: Sorting 4
CSC2100 Data Structure
Yufei Tao
CSE department, CUHK
February 13, 2011
Summary
Problem definition
Counting sort
Summary
In the last lecture, we showed that all algorithms in the comparison
class have a time lower bound of Ω(n log n) (if the array to be
sorted has n numbers). In this lecture, we will learn an algorithm
outside the comparison class. This algorithm, whose name is
counting sort, achieves the cost of O(n) when the data domain has
a size of O(n).
Problem definition
1
Problem definition
2
Counting sort
Main idea
Formal description
Time complexity
Counting sort
Summary
Problem definition
Counting sort
Summary
We first consider a simplified version of the problem that can be
solved by counting sort efficiently. This simplification allows us to
focus on studying the core of the algorithm. It is in fact fairly
straightforward to extend the algorithm to solve the general version
of the problem, as we will see.
Problem (Sort-within-ten)
Let A be an array of n numbers, each of which is an integer from 1
to 10. The objective is to arrange the numbers of A in ascending
order.
Note
Numbers such as 0, -2, 11, and 31 are not allowed in A.
Problem definition
Counting sort
Summary
Main idea
Rationale
Let us first get the idea from an example. Assume that the array
to be sorted is A = {8, 3, 8, 9, 2, 5, 8, 5}.
Example
1
First, create an array cnt of size 10 (i.e., the size of our
domain) with all numbers set to 0.
2
Scan A once. Given each number v read, increase cnt[v ] by 1.
After reading A[1] = 8, increase cnt[8] by 1.
After reading A[2] = 3, increase cnt[3] by 1.
...
At the end of the scan, cnt becomes {0, 1, 1, 0, 2, 0, 0, 3, 1, 0}.
Note
Now cnt[i] (i ≤ 10) equals the number of occurrences of i
in A.
Problem definition
Counting sort
Main idea
Rationale (cont.)
Example (cont.)
3
Create an array pos of size 10, such that pos[i ] (i ≤ 10)
equals the sum of the first i numbers in cnt.
pos = {0, 1, 2, 2, 4, 4, 4, 7, 8, 8}. For example, pos[5] is the
sum of the first 5 numbers in cnt.
Note
pos[i] (i ≤ 10) gives the last position of i in the sorted
order, provided that i belongs to A. For example,
pos[8] = 7 means that, in the sorted order, the last 8 (of
the three 8’s in A) is at the 7-th position.
Summary
Problem definition
Counting sort
Summary
Main idea
Rationale (cont.)
Example (cont.)
4
Create an array A′ of size n for storing the final sorted list.
5
Scan A another time. Given each number v read, copy v to
position pos[v ] in A′ , and decrease pos[v ] by 1 (to indicate
where the next occurrence of v should be).
Put A[1] = 8 at position pos[8] = 7 in A′ ; decrease pos[8] to 6.
Put A[2] = 3 at position pos[3] = 2 in A′ ; decrease pos[3] to 1.
Put A[3] = 8 at position pos[8] = 6 in A′ ; decrease pos[8] to 5.
...
Order of filling A′
{, ,
{ , 3,
{ , 3,
{ , 3,
,
,
,
,
, , , 8, }
, , , 8, }
, , 8, 8, }
, , 8, 8, 9}
...
Problem definition
Counting sort
Formal description
Pseudo-code
Algorithm CSort(A[1..n])
1. cnt = an array of size 10 with all elements set to 0
2. for i = 1 to n
3.
cnt[A[i ]] + +
4. pos = an array of size 10 with all elements set to 0
5. pos[1] = cnt[1]
5. for i = 2 to 10
6.
pos[i ] = pos[i − 1] + cnt[i ]
′
7. A = an array of size n
8. for i = 1 to n
9.
A′ [pos[A[i ]] = A[i ]
10. pos[A[i ]] − −
Summary
Problem definition
Time complexity
Cost analysis
Obviously O(n).
Counting sort
Summary
Problem definition
Counting sort
Summary
Time complexity
General problem
Problem (Small-domain Sort)
Let A be a set of n numbers each of which is an integer from 1 to
k. The objective is to arrange the numbers of A in ascending order.
To think
Extend the counting sort algorithm to solve the above problem in
O(n + k) time.
Problem definition
Counting sort
Playback of this lecture:
Counting sort.
Time complexity O(n + k) where k is the size of the data
domain.
Summary