C950 Data Structures and Algorithms II

Correct Answer

A. for (auto i = c.begin(); i != c.end(); ++i) f(*i);

Explanation

Forward iterators can only move in one direction using the increment operator (++). The comparison i != c.end() is valid for checking if the end has been reached, and using ++i is slightly more efficient than i++ for iterators, especially non-primitive types. Therefore, option A is both correct and most efficient among the valid options.

Why other options are wrong

B. for (auto i = c.begin(); i != c.end(); i++) f(i);

This is incorrect because although it works, using i++ instead of ++i can be slightly less efficient for complex iterators. While functionally the loop behaves correctly, ++i avoids creating a copy of the iterator.

C. for (auto i = c.begin(); i < c.end(); ++i) f(i);

This is incorrect because the < operator is not defined for forward iterators. Forward iterators only support equality/inequality comparisons (== and !=), not relational comparisons like <.

D. for (auto i = c.begin(); i < c.end(); i++) f(i);

This is incorrect for the same reason as option C. Forward iterators do not support the < operator, so this loop will fail to compile or behave incorrectly.

Correct Answer

B. T(n) >= Cg(n)

Explanation

To prove that f(n) = 7n + 3 is Big-Omega(n), we need to show that there exist constants such that the function f(n) is bounded below by a constant multiple of n for sufficiently large n. Specifically, for Big-Omega notation, we assert that T(n) grows at least as fast as cg(n), where c is a constant. Thus, T(n) must be greater than or equal to a constant multiple of g(n).

Why other options are wrong

A. T(n) < Cg(n)

This is incorrect because Big-Omega requires that T(n) be greater than or equal to a constant multiple of g(n), not less than it.

C. T(n) = Cg(n)

This is incorrect because T(n) does not need to be exactly equal to cg(n); it only needs to be greater than or equal to a constant multiple of g(n), as Big-Omega implies a lower bound.

D. T(n) <= Cg(n)

This is incorrect because T(n) should not be bounded above by cg(n) for Big-Omega. The correct relationship is that T(n) is at least a constant multiple of g(n), not less than or equal to it.

Correct Answer

A. Algorithm must use a repetition structure

Explanation

Not all algorithms require a repetition structure (like loops). For example, an algorithm could simply involve a series of steps that are executed once. Repetition (such as loops or recursion) may be a part of the algorithm in some cases, but it is not a necessary requirement for all algorithms.

Why other options are wrong

B. Algorithm must eventually cause a program to terminate

This is a valid requirement for any algorithm. An algorithm must always terminate after a finite number of steps. Algorithms that do not terminate (infinite loops) are not considered valid.

C. Algorithm must give correct solutions for all valid input

This is another valid requirement. An algorithm should provide the correct output for all valid inputs, ensuring the reliability and correctness of the algorithm.

D. Algorithm must be unambiguous

This is an essential requirement. An algorithm must be unambiguous, meaning that each step is clearly defined, leaving no room for interpretation. Ambiguity would result in an unclear or incorrect execution of the algorithm.

Correct Answer

A. fsu::Listfsu::String myList;

Explanation

In C++, to declare a template-based object such as fsu::List that holds elements of type fsu::String, the correct syntax is to specify the type within the angle brackets (<fsu::String>), followed by the object name. Option A is the proper declaration syntax.

Why other options are wrong

B. fsu::Listfsu::String myList = new fsu::Listfsu::String();

This is incorrect because fsu::List is likely a stack-allocated object, and there’s no need for new to dynamically allocate it. It should be declared without new.

C. Listfsu::String myList;

This is incorrect because the class fsu::List is specified with the fsu:: prefix. Without this prefix, it implies a different, possibly undefined class List.

D. fsu::List myListfsu::String;

This is incorrect because the syntax is invalid. The type declaration (fsu::List<fsu::String>) must precede the object name.

Correct Answer

C. Set

Explanation

A set is a collection data structure in which each element must be unique. Sets automatically prevent duplicate elements from being inserted, ensuring that only one instance of each element exists in the collection. This property makes sets different from other collections like lists and arrays, where duplicates can exist.

Why other options are wrong

A. List

A list allows duplicate elements. It is an ordered collection where elements can appear more than once, so it does not ensure uniqueness.

B. Array

An array is a collection of elements, but like a list, it allows duplicates. An array's main characteristic is that it stores elements in a contiguous block of memory and provides fast access to its elements by index, but it does not enforce uniqueness.

D. Dictionary

A dictionary is a collection of key-value pairs, and while each key must be unique, the values can be duplicated. The key is unique, but the values are not necessarily unique, making it different from a set.

Correct Answer

E. Compile

Explanation

"Compile" is not a component of an algorithm. The key components of an algorithm are the steps it takes to solve a problem, which can include instruction (the actions it performs), selection (decisions the algorithm makes), sequence (the order of operations), and repetition (repeating steps until a condition is met). Compilation, on the other hand, refers to the process of converting source code into executable code, which is outside the scope of the algorithm itself.

Why other options are wrong

A. Instruction

Instructions are a core component of an algorithm. They represent the steps or actions the algorithm takes.

B. Selection

Selection is an essential component of algorithms where decisions or conditional branches (like if-else statements) are made.

C. Sequence

Sequence defines the order in which instructions are executed in an algorithm, which is a fundamental part of any algorithm.

D. Repetition

Repetition is also a critical component, as many algorithms involve looping through steps until a condition is satisfied.

Correct Answer

B. Comparison-based sorting algorithms can achieve a worst-case runtime of O(n * log(n)).

Explanation

Comparison-based sorting algorithms, such as mergesort, heapsort, and quicksort, have a worst-case time complexity of O(n * log(n)). This is because these algorithms rely on comparing elements and use divide-and-conquer or similar techniques to efficiently sort the elements. MergeSort, for example, guarantees a worst-case time complexity of O(n * log(n)) regardless of the input order.

Why other options are wrong

A. All comparison-based sorting algorithms have a worst-case runtime of O(n).

This is incorrect. While some algorithms may perform well on certain data sets (like O(n) in best case for quicksort with randomized pivot selection), they do not guarantee this performance in the worst case. A common example is quicksort, which can have a worst-case runtime of O(n^2) if the pivot selection is poor.

C. There exists at least one comparison-based sorting algorithm with a worst-case runtime of O(n^2).

This is correct in the case of algorithms like bubble sort or insertion sort, which have a worst-case runtime of O(n^2). However, the question asks about the relationship between comparison-based sorting algorithms in general, so this statement is true but doesn't apply as universally as B.

D. All comparison-based sorting algorithms have a best-case runtime of Omega(n).

This is incorrect. While comparison-based sorting algorithms do often have a best-case runtime of Omega(n) (e.g., merge sort or heapsort), there are exceptions, like bubble sort or insertion sort, which can perform better under specific conditions (like already sorted input). The best-case runtime is typically O(n) or even Omega(n), but not guaranteed across all algorithms.

Correct Answer

D. Stack

Explanation

A stack is an abstract data type that follows the Last-In-First-Out (LIFO) principle. The two primary operations associated with a stack are push, which adds an element to the top of the stack, and pop, which removes the element from the top. These operations allow elements to be added or removed in a specific order, making stacks useful for problems like function call management, backtracking, and parsing expressions.

Why other options are wrong

A. Queue

A queue is an abstract data type that follows the First-In-First-Out (FIFO) principle. The operations typically associated with a queue are enqueue (to add an element to the end) and dequeue (to remove an element from the front), not push and pop.

B. List

A list is a linear data structure that allows for the storage of elements in a sequence. While lists have operations like insert and delete, they do not specifically define push and pop as the primary methods.

C. Tree

A tree is a hierarchical data structure, and while operations like insert and delete may be used, trees do not typically use the push and pop terminology, which are more associated with stack-based operations.

Correct Answer

A. The root node

Explanation

In preorder traversal, the first node visited is always the root node. The traversal follows the pattern: visit the root, traverse the left subtree, and then traverse the right subtree. The root node is always the starting point of the traversal.

Why other options are wrong

B. The leftmost leaf node

This is incorrect because in preorder traversal, the root node is visited first, not the leftmost leaf. The leftmost node would typically be visited first in an inorder traversal.

C. The rightmost leaf nodeThis is incorrect because the rightmost leaf node is not visited first in preorder traversal. Pre Order starts with the root, and then recursively processes the left and right subtrees.

D. The node with the smallest value

This is incorrect because the smallest value node is not always the first to be visited in preorder traversal. Pre Order starts with the root, regardless of the node values, and then proceeds to the left and right children. The smallest node value would be visited in inorder traversal if the tree is a binary search tree.