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Chapter 1: Problem 6

Write an algorithm that finds both the smallest and largest numbers in a list of \(n\) numbers. Try to find a method that does at most \(1.5 n\) comparisons of array items.

Short Answer

Expert verified
The algorithm iterates through the list by pairing up the numbers, making 3 comparisons for each pair (one within the pair and one each with the current min and max). In case of an odd number of elements, the last unpaired element is compared with the current min and max. By the end of the list, the current min and max are the smallest and largest numbers in the list respectively.

Step by step solution

01

Initialise variables

First, two variables named `min` and `max` need to be initialized, and set to the first element of the array as initial minimal and maximal values respectively.
02

Iterate through the list

Start iterating over the list of numbers from index 1. Pair up the rest elements in the list (i.e., the second with the third, the fourth with the fifth element etc.)
03

Compare pairs and update min and max

First compare the paired elements with each other. The smaller element will then be compared to the current `min` and if it's smaller, it will become the new `min`. The larger element is compared to the current `max` and if it's larger, it will become new `max`. This way, for every two numbers in the list, only 3 comparisons are made.
04

Handle the unpaired element

If the list has an odd length, there will be an single unpaired number at the end of the list. Compare this element with the current `min` and `max` and update `min` and `max` as required.
05

Output the min and max

At the end of the loop, `min` and `max` variables will hold the smallest and largest number in the list respectively.

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Most popular questions from this chapter

There are two algorithms called Algl and Alg2 for a problem of size n. Algl runs in \(n^{2}\) microseconds and Alg2 runs in 100 n log \(n\) microseconds. Algl can be implemented using 4 hours of programmer time and needs 2 minutes of CPU time. On the other hand, Alg2 require 15 hours of programmer time and 6 minutes of CPU time. If programmers are paid 20 dollars per hour and CPU time costs 50 dollars per minute, how many times must a problem instance of size 500 be solved using Alg2 in order to justify its development cost?

Explain in English what functions are in following sets. a. \(n^{O(1)}\) b. \(O\left(n^{O(1)}\right)\) c. \(O\left(O\left(n^{O(1)}\right)\right)\)

Write an algorithm that finds the greatest common divisor of two integers.

Presently we can solve problem instances of size 100 in 1 minute using algorithm \(A,\) which is a \(\Theta\left(2^{n}\right)\) algorithm. On the other hand, we will soon have to solve problem instances twice this large in 1 minute. Do you think it would help to buy a faster (and more expensive) computer?

Show directly that \(f(n)=n^{2}+3 n^{3} \in \Theta\left(n^{3}\right)\). That is, use the definitions of \(O\) and \(\Omega\) to show that \(f(n)\) is in both \(O\left(n^{3)} \text { and } \Omega\left(n^{3)}\right.\right.\)
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