UGC NET CSE | December 2018 | Part 2 | Question: 30

Question

The second smallest of $n$ elements can be found with ____ comparisons in the worst case.

• A. $n-1$
• B. $\lg \: n$
• C. $n + ceil(\lg \: n)-2$
• D. $\frac{3n}{2}$

Comments

https://gateoverflow.in/27194/tifr2014-b-9

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@Shaik Masthan

here ans will be C)

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yes mam, due to we have to add (n-1) comparissions which required to know which is first minimum.