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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

What is the time complexity of the Bellman–Ford algorithm on a graph with V vertices and E edges?

  1. AO(E log V)
  2. BO(V³ × E)
  3. CO(V + E)
  4. DO(V × E)
2.Algorithms & complexityHard

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt can be solved by Dijkstra's algorithm with negated weights
  2. BIt is undecidable
  3. CIt is NP-hard
  4. DIt is always solvable in O(V + E)
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 doubly linked list
  2. BA singly linked list
  3. CA sorted array
  4. DA dynamic array
4.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. AThe need for a hash function
  2. BThe O(n) cost of reading a key's value
  3. CPrimary clustering, where long runs of occupied slots form
  4. DThe need to store keys in the table
5.Graph theoryMedium

A connected graph has 16 vertices and 25 edges. How many edges k must be removed to leave a spanning tree?

|V| = 16|E| = 25
k =
6.Graph theoryHard
10313141251474ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G

3Solutions

1.Algorithms & complexity

Answer: D · O(V × E)

Why this is the answer

Each of the V − 1 rounds relaxes all E edges, giving O(VE); in exchange it handles negative edge weights.

Answer

D · O(V × E)
2.Algorithms & complexity

Answer: C · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

C · It is NP-hard
3.Data structures

Answer: A · 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

A · A doubly linked list
4.Data structures

Answer: C · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

C · Primary clustering, where long runs of occupied slots form
5.Graph theory

Answer: k = 10

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.

Edges of a spanning tree

16 − 1 = 15

Edges to remove

k = 25 − 15 = 10

Answer

k = 10
6.Graph theory

Answer: B

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

1. A (0): B: 10, C: 3
2. C (3): B: 4, G: 17
3. B (4): D: 7, G: 9
4. D (7): F: 21
5. G (9): E: 13
6. E (13)
7. F (21)

round 3

B

Answer

B

ExamProblems · examproblems.com/courses/data-structures-and-algorithms