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

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. AYes, but only if n is smaller than 100
  2. BNo, because n³ is smaller than n² for large n
  3. CYes, because Big O is an upper bound and n² grows no faster than n³
  4. DNo, because Big O must always state the exact growth rate of the function
2.Algorithms & complexityHard

Why can Dijkstra's algorithm give wrong results when some edge weights are negative?

  1. AA vertex is finalised when extracted, but a later negative edge could still give it a shorter path
  2. BNegative numbers cannot be stored in a priority queue
  3. CIt goes into an infinite loop on every negative edge
  4. DIt only works on undirected graphs
3.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA singly linked list
  2. BA doubly linked list
  3. CA sorted array
  4. DA dynamic array
4.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. ADeleted keys must be kept forever for security, so slots can never be reused
  2. BEmptying the slot could break the probe sequence for other keys
  3. CTombstones speed up hashing
  4. DIt makes the table sorted
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
82119212113ABCDEF

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

start: Aend: E
d(E) =

3Solutions

1.Algorithms & complexity

Answer: C · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

C · Yes, because Big O is an upper bound and n² grows no faster than n³
2.Algorithms & complexity

Answer: A · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path

Why this is the answer

The greedy step relies on non-negative weights so that extracted distances are final; Bellman–Ford is used when weights may be negative.

Answer

A · A vertex is finalised when extracted, but a later negative edge could still give it a shorter path
3.Data structures

Answer: B · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

B · A doubly linked list
4.Data structures

Answer: B · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

B · Emptying the slot could break the probe sequence for other keys
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(E) = 19

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: 8
B (8): C: 10, D: 19, E: 21
C (10): E: 19
D (19): F: 31
E (19): F: 20
F (20)

Shortest path

A → B → C → E
d(E) = 19

Answer

d(E) = 19

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