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 multiplying two n × n matrices with the standard triple loop?

  1. AO(n³)
  2. BO(n² log n)
  3. CO(n)
  4. DO(n²)
2.Algorithms & complexityHard

What is the amortised cost of incrementing a binary counter, given that one increment may flip many bits?

  1. AO(n) per increment
  2. BO(log n) per increment
  3. CO(n log n) per increment
  4. DO(1) per increment
3.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AConstant-time search in every case
  2. BThat keys are stored in insertion order
  3. CThat no node has two children
  4. DO(log n) height through rebalancing
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. ANothing changes, because a key's hash is computed only once when the program starts
  2. BThe map automatically moves the key to its new bucket
  3. CLooking up the object may fail because it sits in the bucket for its old hash
  4. DThe map throws an error at the moment the field changes
5.Graph theoryMedium

A tree has 3 vertices of degree 3, 4 vertices of degree 4, 4 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 34 × deg 44 × deg 2
L =
6.Graph theoryHard
10915781410103ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 4; A itself is round 1.

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

3Solutions

1.Algorithms & complexity

Answer: A · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

A · O(n³)
2.Algorithms & complexity

Answer: D · O(1) per increment

Why this is the answer

Over n increments, bit i flips about n / 2ⁱ times, so the total is less than 2n flips, an average of O(1) each.

Answer

D · O(1) per increment
3.Data structures

Answer: D · O(log n) height through rebalancing

Why this is the answer

Self-balancing trees keep height logarithmic, so search, insert and delete are O(log n) in the worst case.

Answer

D · O(log n) height through rebalancing
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: L = 13

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.

Handshake lemma for a tree

L + 33 = 2(L + 11) − 2
L = 13

Answer

L = 13

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

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: 10, C: 9
2. C (9): E: 16, G: 19
3. B (10)
4. E (16): F: 24
5. G (19): D: 29
6. F (24)
7. D (29)

round 4

E

Answer

E

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