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11 votes
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Given a set of $n$ distinct numbers, we would like to determine both the smallest and the largest number. Which of the following statements is TRUE?

  1. These two elements can be determined using $O\left(\log^{100}n\right)$ comparisons.
  2. $O\left(\log^{100}n\right)$ comparisons do not suffice, however these two elements can be determined using $n + O(\log n)$ comparisons.
  3. $n+O(\log n)$ comparisons do not suffice, however these two elements can be determined using $3\lceil n/2 \rceil$ comparisons.
  4. $3\lceil n/2 \rceil$ comparisons do not suffice, however these two elements can be determined using $2(n - 1)$ comparisons.
  5. None of the above.
in Algorithms 2.5k views
1
suppose n = 1024, exact comparisons required will be $1024*1.5 - 2 = 1534$ (C) is the correct option.
0

What is the explanation for it @Manu Thakur.

1
The divide and conquer algorithm for finding the min and max elements of a list operates in worst case $3n/2 - 2$ comparisons. It is described in Sartaj Sahni's algorithms book.
0
Thanks.
2

To find the smallest and the second smallest element (and the third smallest.. so on), it is $n + O(n)$ (See this and this)

To find the maximum and the minimum element, $ceil(3n/2)-2$


2 Answers

12 votes
 
Best answer

I think answer will be C.

To be accurate, it will need $3n/2 -2$ comparisons .


edited by
10

for even numbers- 1.5n-2

for odd- 3/2(n-1) Comparisions

0
@sushmita

Why n+O(logn) doesn't suffice?
0
It'll need exactly

$2(n-1)-\left \lfloor n/2 \right \rfloor = \left\{\begin{matrix} 3\left \lfloor n/2 \right \rfloor & for \; n \in odd \\ 3n/2 -2 & for \; n \in even \end{matrix}\right.$
0
this formula is not correct for n==2
7 votes

Similar to the approach proposed by Himanshu1 here:

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

Construct a decision tree to determine the minimum element: $n - 1$ comparisons

Maximum element can be found from the same tree as it will be the biggest element out of $\frac{n}{2}$ elements at the first level which lost the decision: $\frac{n}{2} - 1$ comparisons

Therefore, the resultant number of comparisons: $3(\frac{n}{2}) - 2$, tighest bound on which is option (C).

0
@pranav Kant Gaur

Maximum element will at leaf na? and  at leaf there are n elements so comparison should be n-1. Please clear my doubt.
0

@Bad_Doctor

Yes maximum comparison will be for leaf.But question is asking for both maximim and minimum .

For maximum n-1 comparison

For minimum n-1 comparison

Total number of comparison=2*(n-1)

If you will draw the tree you will find that the leaf node level(is the last level) needs n/2 comparison.And it is common to both minimum and maximum.

So total number of comparison required=2*(n-1) -n/2 =(3/2)n-2 when n is even.

0

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

This is a rather good explanation.

Answer:

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