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Chapter 1: Problem 2

Write an algorithm that finds the \(m\) smallest numbers in a list of \(n\) numbers.

Short Answer

Expert verified
To find the \(m\) smallest numbers in a list of \(n\) numbers, initialize a list with the first \(m\) numbers, then for each remaining number, if it is smaller than the largest number in your list, replace that number with the current one. Repeat this until you've gone through all the numbers in the list.

Step by step solution

01

Initialize placeholders

Start by initializing a placeholder for the smallest numbers. This will be a list of size \(m\). Initially, you can fill it with the first \(m\) elements of the list.
02

Scan the list

Now scan through the list starting from index \(m\). For each element, compare it with the largest element in your smallest numbers list (which can be efficiently found if we keep the list sorted or use a max heap data structure).
03

Update placeholders

If the current element is smaller than the largest element in your 'smallest numbers' list, replace that largest element with the current one. Make sure to keep the 'smallest numbers' list sorted or the heap property maintained after this operation.
04

Iterate through the list

Repeat Step 3 for all elements in the list. At the end, your 'smallest numbers' list or min heap will contain the \(m\) smallest elements of the original list.

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Most popular questions from this chapter

Algorithm 1.7 (nth Fibonacci Term, Iterative) is clearly linear in \(n,\) but is it a linear-time algorithm? In Section 1.3 .1 we defined the input size as the size of the input. In the case of the \(nth\) Fibonacci term, \(n\) is the input, and the number of bits it takes to encode \(n\) could be used as the input size. Using this measure the size of 64 is \(\lg 64=6,\) and the size of 1024 is \(\lg 1024=\) 10\. Show that Algorithm 1.7 is exponential-time in terms of its input size. Show further that any algorithm for computing the \(nth\) Fibonacci term must be an exponential-time algorithm because the size of the output is exponential in the input size. See Section 9.2 for a related discussion of the input size.

Let \(p(n)=a_{4} n^{4}+a_{k-1} n^{k-1}+\dots a_{1} n+a_{0},\) where \(a_{k}>0 .\) Using the Properties of Order in Section \(1.4 .2,\) show that \(p(n) \in \Theta\left(n^{k}\right)\)

Discuss the reflexive, symmetric, and transitive properties for asymptotic comparisons \((O, \Omega, \theta, o)\)

What is the time complexity \(T(n)\) of the nested loops below? For simplicity, you may assume that \(n\) is a power of \(2 .\) That is, \(n=2^{k}\) for some positive integer \(k\) : \\[ \begin{array}{l} i=n_{i} \\ \text { while }(i>=1)\\{ \\ \qquad \begin{array}{c} j=i \\ \text { while }(j<=n) \end{array} \end{array} \\] < body of the inner while loop> \(\quad\) I/ Needs \(\Theta(1)\) \\[ j=2^{*} j \\] } \\[ i=|1 / 2| \\] }

Give an algorithm for the following problem. Given a list of \(n\) distinct positive integers, partition the list into two sublists, each of size \(n / 2,\) such that the difference between the sums of the integers in the two sublists is minimized. Determine the time complexity of your algorithm. You may assume that \(n\) is a multiple of 2
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