Data Structures and Algorithms: cram pack

Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.

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1Formulas and methods

Graph theory

Handshake lemma
Σ deg(v) = 2|E|
Complete graphs
|E(Kn)| = n(n − 1)/2, |E(Km,n)| = m·n
Regular graph
|E| = n·k/2
Tree and forest
|E| = n − 1, |E| = n − c
Leaves of a tree
L = 2 + Σ (deg(v) − 2) over deg(v) ≥ 3
Euler
circuit: all degrees even; path: exactly 2 odd degrees
Hamilton circuits in Kn
(n − 1)!/2
Dijkstra update
d(v) = min(d(v), d(u) + w(u, v))
Walks of length L
(Mᴸ)ij

Method

  1. Write down the degree of every vertex; the degrees add up to twice the number of edges.
  2. For Euler questions, check that the graph is connected and count the vertices with odd degree.
  3. For trees, use n − 1 edges and the handshake lemma.
  4. For Dijkstra, keep a table of tentative distances and settle the smallest unsettled vertex each round.
  5. For a minimum spanning tree, sort the edges by weight and skip every edge that closes a cycle (Kruskal), or grow one tree from a start vertex (Prim).

2Practice problems

1.Algorithms & complexityMedium

The lower bound for comparison-based sorting in the worst case is:

  1. AΩ(log n)
  2. BΩ(n log n)
  3. CΩ(n)
  4. DΩ(n²)
2.Algorithms & complexityHard

Strassen's algorithm multiplies matrices using 7 instead of 8 recursive multiplications of half-size blocks. What is its running time?

  1. AΘ(n³)
  2. BΘ(n² log n)
  3. CΘ(n2)
  4. DΘ(nlog₂ 7), about n2.81
3.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ADouble hashing
  2. BLinear probing
  3. CQuadratic probing
  4. DSeparate chaining
4.Data structuresHard

Kahn's algorithm for topological sorting repeatedly removes vertices with in-degree 0. Which structure holds those vertices?

  1. AA stack of all visited edges
  2. BA max-heap ordered by out-degree
  3. CA queue (or any collection) of vertices whose in-degree is 0
  4. DAn adjacency matrix recording which vertices have already been removed
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
6.Graph theoryHard
1251011910312ABCDEF

Use Dijkstra’s algorithm to find the length d(C) of the shortest path from A to C in the weighted graph.

start: Aend: C
d(C) =

3Solutions

1.Algorithms & complexity

Answer: B · Ω(n log n)

Why this is the answer

Any comparison sort needs at least log₂(n!) ≈ n log n comparisons in the worst case. Counting sort beats it because it does not compare.

Answer

B · Ω(n log n)
2.Algorithms & complexity

Answer: D · Θ(nlog₂ 7), about n2.81

Why this is the answer

With a = 7 and b = 2, nlog₂ 7 dominates the O(n²) addition work, giving about n2.81.

Answer

D · Θ(nlog₂ 7), about n2.81
3.Data structures

Answer: D · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

D · Separate chaining
4.Data structures

Answer: C · A queue (or any collection) of vertices whose in-degree is 0

Why this is the answer

Ready vertices are kept in a queue; removing one lowers its neighbours' in-degrees, and any that reach 0 are added, giving O(V + E) time.

Answer

C · A queue (or any collection) of vertices whose in-degree is 0
5.Graph theory

Answer: H = 3

Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.

Count

H = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
6.Graph theory

Answer: d(C) = 17

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Dijkstra round by round

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 12
B (12): C: 17, D: 22, E: 24
C (17)
D (22): F: 32
E (24): F: 27
F (27)

Shortest path

A → B → C
d(C) = 17

Answer

d(C) = 17

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