COUNTING SORT
Design & Analysis of Algorithms
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Counting Sort-Sorting Algorithm Design and Analysis-Lecture Slides, Slides of Design and Analysis of Algorithms
This lecture is part of lecture series for Design and Analysis of Algorithms course. This course was taught by Dr. Bhaskar Sanyal at Maulana Azad National Institute of Technology. It includes: Design, Analysis, Algorithms, Counting, Sort, Linear, Time, Auxiliary, Storage, Loop, Running
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2011/2012
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COUNTING SORT
Design & Analysis of Algorithms
Sorting in linear time
Counting sort: No comparisons between elements.
• Input : A [1.. n ], where A [ j ]{1, 2, …, k }.
• Output : B [1.. n ], sorted.
• Auxiliary storage : C [1.. k ].
Counting-sort example
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 2 3 4
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 0 0 0 0 1 2 3 4 for i 1 to k do C [ i ] 0
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 0 0 1 1 2 3 4 for j 1 to n do C [ A [ j ]] C [ A [ j ]] + 1 ⊳ C [ i ] = |{key = i }|
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 0 1 1 1 2 3 4 for j 1 to n do C [ A [ j ]] C [ A [ j ]] + 1 ⊳ C [ i ] = |{key = i }|
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 0 2 2 1 2 3 4 for j 1 to n do C [ A [ j ]] C [ A [ j ]] + 1 ⊳ C [ i ] = |{key = i }|
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 0 2 2 1 2 3 4 C' : 1 1 2 2 for i 2 to k do C [ i ] C [ i ] + C [ i – 1] ⊳ C [ i ] = |{key i }|
A : 4 1 3 4 3
B :
1 2 3 4 5 C : 1 0 2 2 1 2 3 4 C' : 1 1 3 5 for i 2 to k do C [ i ] C [ i ] + C [ i – 1] ⊳ C [ i ] = |{key i }|
A : 4 1 3 4 3
B : 3
1 2 3 4 5 C : 1 1 3 5 1 2 3 4 C' : 1 1 2 5 for j n downto 1 do B [ C [ A [ j ]]] A[ j ] C [ A [ j ]] C [ A [ j ]] – 1
A : 4 1 3 4 3
B : 3 3 4
1 2 3 4 5 C : 1 1 2 4 1 2 3 4 C' : 1 1 1 4 for j n downto 1 do B [ C [ A [ j ]]] A[ j ] C [ A [ j ]] C [ A [ j ]] – 1
A : 4 1 3 4 3
B : 1 3 3 4
1 2 3 4 5 C : 1 1 1 4 1 2 3 4 C' : 0 1 1 4 for j n downto 1 do B [ C [ A [ j ]]] A[ j ] C [ A [ j ]] C [ A [ j ]] – 1
Analysis
for i 1 to k do C [ i ] 0 ( n ) ( k ) ( n ) ( k ) for j 1 to n do C [ A [ j ]] C [ A [ j ]] + 1 for i 2 to k do C [ i ] C [ i ] + C [ i – 1] for j n downto 1 do B [ C [ A [ j ]]] A[ j ] C [ A [ j ]] C [ A [ j ]] – 1 ( n + k )
Running time
If k = O ( n ), then counting sort takes ( n ) time.
• But, sorting takes ( n lg n ) time!
• Where’s the fallacy?
Answer:
• Comparison sorting takes ( n lg n ) time.
• Counting sort is not a comparison sort.
• In fact, not a single comparison between
elements occurs!