madeeasy

Question

Answer 1

Take 3 temporary elements ...

Max1 , max2, max3

Now make 0th element  as max1 check with all elements when when max1 changes put previous value of max1 into max2 ...

Now second time max1 changes that means new greater element has been found...

Puprevious max1 into max 2 and previous max2 value into max3...

Now this process repeated till the end of array...

Which take O(n) time...

Comments

Not comparing sir...just shifting..

When max1 changes with 1 comparison than just shifting...

Old value Max2 to max3 and old value max1 to max2 and new maximum into max1...

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7 5 6 2 3 4

will it work here?

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Yes..I got it...

When max1 will not change ....even there is new second largest available my approach can't determine it...Thanks... @Arjun sir

@kapil

Answer 2

Answer will be C)

The list is not sorted.

So to get max. of any 3 numbers we need 2 comparison.

Now, to get max. of n numbers we need 2n/3 comparison.

So, total maximum comparison O(n)

Comments

n and O(n) are not same