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During
the first pass, there are N-1
comparisons
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and
at most N-1 exchanges.
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During
the second pass, there are N-2 comparisons
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and
at most N-2 exchanges.
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Therefore,
in the worst case there are comparisons
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of:
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(N-1)+(N-2)+...+ 1 = N*(N-1)/2
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and,
the same number of exchanges... N*(N-1)/2*4
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which is: 2N*N
- 2*N
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