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

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

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

Which of these satisfiability problems is known to be solvable in polynomial time?

  1. A2-SAT, where each clause has at most two literals
  2. BGeneral SAT on formulas in conjunctive normal form
  3. CCircuit satisfiability
  4. D3-SAT, where each clause has exactly three literals
3.Data structuresMedium

What is the time complexity of searching for a key in a balanced binary search tree?

  1. AO(log n)
  2. BO(n)
  3. CO(1)
  4. DO(n log n)
4.Data structuresHard

An object is stored as a key in a hash map, and then a field used by its hash function is changed. What is the likely result?

  1. AThe map throws an error at the moment the field changes
  2. BNothing changes, because a key's hash is computed only once when the program starts
  3. CLooking up the object may fail because it sits in the bucket for its old hash
  4. DThe map automatically moves the key to its new bucket
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
513147141441512ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 5; A itself is round 1.

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

3Solutions

1.Algorithms & complexity

Answer: C · 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

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

Answer: A · 2-SAT, where each clause has at most two literals

Why this is the answer

2-SAT is in P (linear time via strongly connected components), while 3-SAT is NP-complete.

Answer

A · 2-SAT, where each clause has at most two literals
3.Data structures

Answer: A · O(log n)

Why this is the answer

A balanced tree has height about log₂ n, so a search follows one path from root to leaf.

Answer

A · O(log n)
4.Data structures

Answer: C · Looking up the object may fail because it sits in the bucket for its old hash

Why this is the answer

The key's location was chosen from its old hash; with a new hash, lookups search the wrong bucket, so keys should be immutable.

Answer

C · Looking up the object may fail because it sits in the bucket for its old hash
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: E

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: 5, C: 13
2. B (5): G: 19
3. C (13): G: 17
4. G (17): D: 32, E: 29
5. E (29)
6. D (32): F: 46
7. F (46)

round 5

E

Answer

E

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