ace series

Question

 

If we use bubble sort and run the outer loop for 3 rounds we can get 3 smallest elements  in at most (n-1+n-2+n-3) i.e. 3n-6 comparisons

in that way (C) is also true

Comments

Yup

but answer given is A) right ??

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ans given is actually C using heap and delete from heap operation

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hmm that's also a nice approach

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O(n)