Bucketsort

Bucket Sort

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General information

Algorithmic problem: Sorting sequences of strings

Type of algorithm: loop

Axiliary data:

1. An ordered sequence [math]\displaystyle{ S' }[/math] of strings, which will eventually hold the overall result of the algorithm.
2. The buckets, that is, an array [math]\displaystyle{ B }[/math] whose index range contains the ID range of [math]\displaystyle{ \Sigma }[/math] (e.g. 26 for the alphabet) and whose components are ordered sequences of strings.
3. Let [math]\displaystyle{ N }[/math] denote the maximum length of an input string.
4. An array [math]\displaystyle{ A }[/math] with index range [math]\displaystyle{ [1,\dots,N] }[/math] and multisets of strings as components.

Abstract view

Invariant: After [math]\displaystyle{ i \geq 0 }[/math] iterations:

1. For [math]\displaystyle{ j \in \{1,\dots,N-i\} }[/math], [math]\displaystyle{ A[j] }[/math] contains all input strings of length [math]\displaystyle{ j }[/math].
2. [math]\displaystyle{ S' }[/math] contains all other input strings, that is, the ones with length at least [math]\displaystyle{ N - i + 1 }[/math]. The sequence [math]\displaystyle{ S' }[/math] is sorted according to the following definition of comparison: [math]\displaystyle{ str1 \lt str2 }[/math] (resp. [math]\displaystyle{ str1 \leq str2 }[/math]) means that the substring of [math]\displaystyle{ str1 }[/math] starting at position [math]\displaystyle{ N - i + 1 }[/math] is lexicographically smaller (resp., smaller or equal) than the substring of [math]\displaystyle{ str2 }[/math] starting at position [math]\displaystyle{ N - i + 1 }[/math].
3. All buckets are empty.

Variant: [math]\displaystyle{ i }[/math] increases by [math]\displaystyle{ 1 }[/math]

Break condition: [math]\displaystyle{ i = N }[/math].

Complexity

Statement: Let [math]\displaystyle{ M }[/math] denote the total sum of all input string lengths. Then the asymptotic complexity is in [math]\displaystyle{ \Theta(M) }[/math] in the best and the worst case.

Proof: Obviously, the preprocessing takes <mathO(M)</math> time. In the main loop, each character of each string is read exactly once. Obviously, no operation is applied more often than the reading of single characters.