The time complexity of counting sort

Question

I am taking an algorithms course and there I saw that the time complexity of counting sort is O(n+k) where k is the range of numbers and n is the input size. My question is, when the difference between k and n is too much, such as when k=O(n²)or O(n³), can we say that the complexity is O(n²) or O(n³)? Then in this case counting sort is not a wise approach and we can use merge sort. Am I right?

NPE · Accepted Answer · 2013-03-24T14:37:20.263

10

Yes, you are exactly right on all counts.

Furthermore, we can make stronger statements: when k=O(n²) or O(n³), we can say that the complexity of the counting sort is Θ(n²) or Θ(n³).

edited Mar 24 '13 at 14:37

answered Mar 24 '13 at 14:24

NPE

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Thanks for the answer, i just wanted to make sure whether i understood right – Mar 24 '13 at 14:25

score 3 · Answer 2 · answered Jul 09 '13 at 10:04

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You can still sort in O(n) time, theoretically. If the range of values is say 1 to n³ then convert to base n and do a Radix sort. In base n the number has 3 digits, so your run time is O(3n + 3n) + O(n) for the base conversion. Overall O(n).

answered Jul 09 '13 at 10:04

Robert S. Barnes

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you are assuming `O(1)` operations on base `n` numbers. It seems unlikely to hold in general. – jfs Jan 06 '14 at 23:14

The time complexity of counting sort

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