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

You sort an array of n elements once and then answer k membership queries with binary search. What is the total cost?

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

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

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

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n), because later elements must shift
  2. BO(log n), because of binary search
  3. CO(n²), because each element moves n times
  4. DO(1), because the index is known
4.Data structuresHard

Why might a server choose its hash function at random from a universal family?

  1. ATo make the hash table sorted
  2. BTo guarantee that no two keys can ever collide, whatever keys an attacker sends
  3. CTo make it very unlikely that an attacker can pick many keys that all collide
  4. DTo remove the need to resize the table
5.Graph theoryMedium

A connected graph has 17 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?

|V| = 17|E| = 18
k =
6.Graph theoryHard
13111220981510ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 5.

start: Dk = 5
  1. 1edge AB
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE

3Solutions

1.Algorithms & complexity

Answer: A · O(n log n + k log n)

Why this is the answer

Sorting costs O(n log n) and each query O(log n); this beats k linear scans (O(nk)) when k is large.

Answer

A · O(n log n + k log n)
2.Algorithms & complexity

Answer: D · 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

D · Θ(n log² n)
3.Data structures

Answer: A · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

A · O(n), because later elements must shift
4.Data structures

Answer: C · To make it very unlikely that an attacker can pick many keys that all collide

Why this is the answer

With universal hashing the expected number of collisions is small for any set of keys, since the attacker does not know which function was chosen.

Answer

C · To make it very unlikely that an attacker can pick many keys that all collide
5.Graph theory

Answer: k = 2

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

17 − 1 = 16

Edges to remove

k = 18 − 16 = 2

Answer

k = 2
6.Graph theory

Answer: edge AB

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.

Prim’s algorithm

1. DF (8)
2. DE (9)
3. BE (10)
4. BC (11)
5. AB (13)

edge number 5

AB

Answer

edge AB

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