WGU C949 DATA STRUCTURES AND ALGORITHMS A MASTERY 150 CORRECT VERIFIED ANSWERS WITH DETAILED RATIONALES INSTANT DOWNLOAD

WGU C949 DATA STRUCTURES
AND ALGORITHMS A MASTERY
150 CORRECT VERIFIED ANSWERS
WITH DETAILED RATIONALES
INSTANT DOWNLOAD

WGU C949 DATA STRUCTURES AND ALGORITHMS
Comprehensive Study Guide with Detailed Rationales
Question 1: A software engineer is analyzing the performance of a
newly developed search algorithm designed to find a specific integer
within an unsorted array of N elements; the engineer observes that in the
worst-case scenario, the algorithm must examine every single element in
the array before concluding that the target is not present. Based on this
observation, what is the worst-case time complexity of this algorithm
expressed in Big O notation? A) O(1) B) O(log N) C) O(N) D) O(N log N)
Correct Answer: C - O(N) Rationale: In an unsorted array, the only
way to guarantee finding an element (or determining its absence) is to
check each element one by one. In the worst case, the target is either the
last element or not in the array at all, requiring exactly N comparisons.
Therefore, the worst-case time complexity is linear, denoted as O(N).


Question 2: When evaluating the space complexity of a recursive
algorithm that calculates the factorial of a number N, a computer science
student notes that each recursive call adds a new frame to the call stack;
considering that the recursion depth reaches N before hitting the base
case, what is the auxiliary space complexity of this recursive factorial
function? A) O(1) B) O(log N) C) O(N) D) O(N^2) Correct Answer: C -
O(N) Rationale: Space complexity measures the amount of extra
memory an algorithm needs relative to the input size. In a recursive
function, each call consumes space on the call stack to store local
variables and the return address. Since the factorial function recurses
N times before reaching the base case, there will be N frames on the call
stack simultaneously, resulting in an auxiliary space complexity of
O(N).

,Question 3: A developer notices that as the input size of a specific
sorting algorithm doubles, the execution time increases by a factor of
four; assuming the algorithm's performance is strictly governed by its
dominant term, which of the following time complexities best describes
this algorithm's behavior? A) O(N) B) O(N log N) C) O(N^2) D) O(2^N)
Correct Answer: C - O(N^2) Rationale: If doubling the input size
(N) results in the execution time increasing by a factor of four (2^2), the
relationship between input size and time is quadratic. Mathematically,
if T(N) = c * N^2, then T(2N) = c * (2N)^2 = 4 * c * N^2 = 4 * T(N). This
quadratic growth rate is characteristic of O(N^2) algorithms, such as
Bubble Sort or Insertion Sort.


Question 4: In the context of algorithm analysis, a student is reviewing
the performance of a linear search algorithm on a sorted array of 1,000
elements; the student is asked to identify the best-case time complexity
for this specific scenario. What is the best-case time complexity, and
under what condition does it occur? A) O(1), when the target element is
the very first element in the array. B) O(1), when the target element is
located exactly in the middle of the array. C) O(log N), when the target
element is found using a binary search approach. D) O(N), when the
target element is the last element in the array. Correct Answer: A -
O(1), when the target element is the very first element in the
array. Rationale: The best-case time complexity represents the
minimum amount of time an algorithm will take given the most
favorable input. For a linear search, the most favorable scenario is
when the target element is located at the first index (index 0). The
algorithm checks the first element, finds a match, and terminates
immediately, requiring only one comparison, which is O(1) or constant
time.


Question 5: In algorithm analysis, what does the "O" in Big O notation
specifically represent when we describe an algorithm's time complexity
as O(f(N))? A) The exact number of operations the algorithm will
perform for an input of size N. B) The lower bound of the algorithm's
running time, representing the best-case scenario. C) The upper bound
of the algorithm's running time, representing the worst-case or
asymptotic growth rate. D) The average number of operations the

,algorithm performs across all possible inputs of size N. Correct
Answer: C - The upper bound of the algorithm's running time,
representing the worst-case or asymptotic growth rate.
Rationale: Big O notation specifically describes the asymptotic upper
bound of an algorithm's growth rate. It provides a worst-case
guarantee on how the runtime or space requirements will scale as the
input size (N) grows toward infinity, ignoring constant factors and
lower-order terms.


Question 6: A system needs to store a fixed number of sensor readings
collected every millisecond for exactly one hour; the development team
decides to use a standard static array to store these readings because the
total number of readings is known in advance and will not change. What
is the primary advantage of using a static array in this specific scenario?
A) It allows for dynamic resizing as the number of readings fluctuates. B)
It provides O(1) constant time complexity for accessing any element by
its index. C) It automatically sorts the sensor readings as they are
inserted. D) It prevents buffer overflow errors by dynamically allocating
memory from the heap. Correct Answer: B - It provides O(1)
constant time complexity for accessing any element by its
index. Rationale: The primary advantage of a static array is that its
elements are stored in contiguous memory locations. This allows for
direct, random access to any element using its index in O(1) constant
time, calculated via the base address plus the index multiplied by the
element size. Since the size is fixed and known, a static array is highly
efficient and avoids the overhead of dynamic resizing.


Question 7: You are tasked with implementing a dynamic array (like an
ArrayList in Java or std::vector in C++) that automatically resizes itself
when it runs out of capacity; when the array becomes full, you decide to
double its current capacity and copy all existing elements to the new,
larger array. What is the amortized time complexity of inserting a new
element into this dynamic array? A) O(1) B) O(log N) C) O(N) D) O(N log
N) Correct Answer: A - O(1) Rationale: While resizing the array
takes O(N) time because all elements must be copied, resizing happens
infrequently (only when the array is full). If the array doubles in size
each time, the cost of copying is spread out (amortized) over the N
insertions that occurred since the last resize. Therefore, the amortized

, time complexity for an insertion is O(1), even though the worst-case
time for a single insertion that triggers a resize is O(N).


Question 8: A matrix multiplication algorithm requires iterating
through a two-dimensional array to compute the dot products of rows
and columns; if the algorithm uses three nested loops to multiply an N x
N matrix by another N x N matrix, what is the overall time complexity of
this standard matrix multiplication algorithm? A) O(N) B) O(N^2) C)
O(N^3) D) O(N log N) Correct Answer: C - O(N^3) Rationale:
Standard matrix multiplication involves computing each element of the
resulting N x N matrix. There are N^2 elements in the result matrix. To
compute each element, the algorithm must perform a dot product of a
row and a column, which requires N multiplications and additions.
Therefore, the total number of operations is N^2 * N = N^3, resulting in
a time complexity of O(N^3).


Question 9: In a zero-indexed array of size N, where each element
occupies K bytes of memory, what is the correct formula to calculate the
memory address of the element at index i, given that the base address of
the array is B? A) Address = B + (i * K) B) Address = B + (i * N) C)
Address = B + (K * N) + i D) Address = B + (i / K) Correct Answer: A -
Address = B + (i * K) Rationale: Because arrays store elements in
contiguous memory locations, the address of any element can be
calculated directly. The memory address of the element at index i is
found by taking the base address of the array (B) and adding the offset.
The offset is the index (i) multiplied by the size of each element in bytes
(K). Therefore, Address = B + (i * K).


Question 10: A developer is choosing between an array and a singly
linked list for an application that requires frequent insertions and
deletions in the middle of the data structure, but very few random
accesses; which data structure is more efficient for the insertion and
deletion operations, and why? A) Array, because it allows O(1) insertion
and deletion at any index. B) Singly linked list, because it allows O(1)
insertion and deletion once the correct position is found, without shifting
elements. C) Array, because it automatically shifts elements in O(1) time.
D) Singly linked list, because it allows O(1) random access to find the