minimum number of comparisons

Question

You are given an array of 64 elements, minimum number of comparisons required to find out second largest element among all will be _______.

Comments

what Explaination they given.

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no explanation given.

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Plz go through this :

https://stackoverflow.com/questions/3628718/find-the-2nd-largest-element-in-an-array-with-minimum-number-of-comparisons

63 comparison is required to find the first maximum element itself..For rest of the explanation , plz refer to the above link @Anu007 and @shaurya vardhan..

Answer 1

To find the second largest element, we need to find largest element first.

Number of comparisons to find largest number = n - 1 = 64 - 1 = 63

Now, the second largest must be in the losers of first largest i.e. it might have descended down logn levels at most.

So, number of comparisons to find second largest among logn elements = logn - 1 = log(64) - 1 = 6 - 1 = 5

So total comparisons in all = (n -1) + (logn - 1) = 63 + 5 = 68

Comments

But @just_bhavana for your answer to be correct.....it should be mentioned that extra space is given

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@Anurag your approach is wrong.....think what if 7 was at the starting compared to 8, then it would be at the lowest level

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got it THANKS