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WGU C949 Final Exam Data Structures and Algorithms – Higher Education Computer Science Curriculum – 2026/2027 Academic Year – Verified Questions and Answers for Advanced Data Structures and Algorithms Learners

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This document contains verified questions and answers for the WGU C949 Final Exam Data Structures and Algorithms for the 2026/2027 academic year. It covers core computer science concepts, including arrays, linked lists, stacks, queues, trees, graphs, hash tables, sorting and searching algorithms, recursion, algorithm analysis, Big O notation, and problem-solving techniques. This comprehensive 50-question exam set is designed to reinforce data structures and algorithms knowledge and support preparation for the WGU C949 final assessment.

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WGU C949 FINAL EXAM DATA
STRUCTURES AND ALGORITHMS C949
WGU FINAL EXAM 2026/2027 | Verified
Questions
Higher Education Computer Science Curriculum | Verified Q&A
| Advanced Data Structures and Algorithms Learners
Comprehensive 50-Question Exam Set | 2026/2027


Introduction
This original 2026/2027 comprehensive 50-question set covers Algorithmic Analysis and Complexity
Reasoning, Core Data Structures and Abstract Data Types, Algorithm Design, Sorting, Searching, and
Optimization, and Trees, Graphs, and Advanced Data Relationship Reasoning. The document contains 50
questions designed to reinforce official WGU C949 course objectives for actual exam readiness and advanced
computational proficiency while strengthening higher education mastery of asymptotic analysis,
implementation tradeoffs, algorithm correctness, data abstraction, and graph/tree reasoning. Each question is
written as original practice content with a focused rationale that supports graduate-level and professional
preparation for rigorous data structures and algorithms assessment.




Actual Questions
Domain: Algorithmic Analysis and Complexity Reasoning
1. When analyzing an algorithm whose running time is 7n log n + 40n + 900, which
classification best expresses its asymptotic growth?

A. O(n)

B. O(n log n)

C. O(n^2)

D. O(log n)

Correct Answer: B

Rationale: The dominant term n log n determines the asymptotic class because constants and lower-order
terms become insignificant as input size grows. This matches WGU C949 complexity-reasoning objectives
for comparing algorithms by growth rate rather than machine-specific timing.

2. For the recurrence T(n) = 3T(n/2) + n, which Big-Theta bound is obtained using standard
divide-and-conquer recurrence analysis?

A. Theta(n)


WGU C949 FINAL EXAM DATA STRUCTURES AND ALGORITHMS C949 WGU FINAL EXAM 2026/2027 | Verified Questions

, B. Theta(n log n)

C. Theta(n^2)

D. Theta(n^log2(3))

Correct Answer: D

Rationale: The recursive work has a = 3 subproblems of size n/2, so n^log2(3) dominates the linear combine
step. C949 analysis emphasizes identifying the recursion tree or Master Theorem case that controls total
growth.

3. A dynamic array doubles its capacity whenever it becomes full. What is the amortized time
complexity of appending one item over a long sequence of appends?

A. O(1)

B. O(log n)

C. O(n)

D. O(n log n)

Correct Answer: A

Rationale: Although an individual resize can copy many elements, the total copying cost spreads across
many constant-time appends, yielding amortized O(1). This is a core C949 example of aggregate analysis.

4. Which statement best captures the comparison-sort lower bound for sorting arbitrary
comparable keys?

A. Any comparison sort can always sort in linear time.

B. The lower bound applies only to recursive sorts.

C. Any comparison sort requires Omega(n log n) comparisons in the worst case.

D. The lower bound is Omega(log n) because each comparison is binary.

Correct Answer: C

Rationale: The decision-tree model requires enough leaves to represent all n! input permutations, producing
an Omega(n log n) lower bound. C949 sorting analysis uses this bound to distinguish comparison and non-
comparison methods.

5. A hash-table search is O(1) on average under a good hash function and controlled load
factor, but O(n) in the worst case. What reasoning explains this difference?

A. Average case ignores collisions entirely.

B. A bad collision pattern can place many keys in one probe sequence or bucket.

C. Hash tables always use sorted arrays internally.

D. Worst-case time is O(log n) because hashing is balanced.

Correct Answer: B

Rationale: Average behavior assumes reasonably uniform distribution, while worst-case analysis must
allow all keys to collide. C949 performance reasoning separates expected behavior from adversarial or
degenerate inputs.

WGU C949 FINAL EXAM DATA STRUCTURES AND ALGORITHMS C949 WGU FINAL EXAM 2026/2027 | Verified Questions

, 6. Which scenario most directly illustrates a space-time tradeoff?

A. Storing memoized subproblem results to avoid recomputing them.

B. Replacing all loops with recursion without storing additional state.

C. Deleting indices to minimize memory while increasing scans.

D. Using a slower disk instead of faster memory.

Correct Answer: A

Rationale: Memoization spends additional memory on cached results so repeated subproblems can be
answered faster. This aligns with advanced algorithm-design objectives that evaluate memory cost
alongside running time.

7. If f(n) is Theta(g(n)), what relationship must hold between f and g for sufficiently large n?

A. f eventually becomes exactly equal to g.

B. f must be smaller than g for every input.

C. g must be a polynomial function.

D. f is bounded both above and below by constant multiples of g.

Correct Answer: D

Rationale: Theta notation gives a tight asymptotic bound, requiring both an upper and lower constant-
bound relationship after a threshold input size. C949 uses this language to make precise complexity claims.

8. An algorithm repeatedly halves the remaining search interval until one item remains.
Which complexity class best describes the number of iterations?

A. O(n)

B. O(n/2)

C. O(log n)

D. O(n log n)

Correct Answer: C

Rationale: Halving produces a logarithmic number of steps because k halvings reduce n to 1 when k is
approximately log2(n). This is a standard C949 model for binary-search-style reasoning.

9. A loop executes for i = 1 to n, and an inner loop executes exactly i times for each i. What is
the asymptotic total number of inner-loop executions?

A. Theta(n^2)

B. Theta(n log n)

C. Theta(log n)

D. Theta(n)

Correct Answer: A

Rationale: The total work is 1 + 2 + ... + n, which equals n(n + 1)/2 and therefore grows quadratically. C949
expects learners to translate loop structure into summations.
WGU C949 FINAL EXAM DATA STRUCTURES AND ALGORITHMS C949 WGU FINAL EXAM 2026/2027 | Verified Questions

Información del documento

Subido en
6 de agosto de 2026
Número de páginas
17
Escrito en
2026/2027
Tipo
Examen
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