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 worst-case time complexity of quicksort?

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

What does the recurrence T(n) = 2T(n/2) + n log n solve to?

  1. AΘ(n)
  2. BΘ(n log² n)
  3. CΘ(n²)
  4. DΘ(n log n)
3.Data structuresMedium

AVL trees and red-black trees guarantee:

  1. AO(log n) height through rebalancing
  2. BThat keys are stored in insertion order
  3. CThat no node has two children
  4. DConstant-time search in every case
4.Data structuresHard

Why do databases often use B+ trees rather than plain B-trees for indexes?

  1. AAll records are in linked leaves, which makes range scans efficient
  2. BB+ trees use binary internal nodes, which reduces the number of disk reads
  3. CB+ trees do not need to be balanced
  4. DB+ trees store data only in the root
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
111151521073ABCDEF

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

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: B · O(n²)

Why this is the answer

With consistently bad pivots (e.g. already sorted input and first-element pivot) partitions are unbalanced, giving O(n²). The average case is O(n log n).

Answer

B · O(n²)
2.Algorithms & complexity

Answer: B · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

B · Θ(n log² n)
3.Data structures

Answer: A · 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

A · O(log n) height through rebalancing
4.Data structures

Answer: A · All records are in linked leaves, which makes range scans efficient

Why this is the answer

In a B+ tree internal nodes only guide the search and the leaves are linked in key order, so a range query finds the first key and then walks along the leaves.

Answer

A · All records are in linked leaves, which makes range scans efficient
5.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
6.Graph theory

Answer: d(F) = 21

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: 1, C: 11
B (1): D: 16, E: 4
E (4): C: 6, D: 14
C (6)
D (14): F: 21
F (21)

Shortest path

A → B → E → D → F
d(F) = 21

Answer

d(F) = 21

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