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 does the recurrence T(n) = 2T(n − 1) + O(1) solve to?

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

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n²)
  2. BΘ(n² log n)
  3. CΘ(n³)
  4. DΘ(n log n)
3.Data structuresMedium

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

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

How long does it take to search for an arbitrary key in a binary min-heap of n elements?

  1. AO(1)
  2. BO(log n)
  3. CO(n)
  4. DO(log log n)
5.Graph theoryMedium
ABCDEEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 5|E| = 7
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
12771312121011ABCDEF

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

3Solutions

1.Algorithms & complexity

Answer: B · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

B · O(2ⁿ)
2.Algorithms & complexity

Answer: C · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

C · Θ(n³)
3.Data structures

Answer: D · 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

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

Answer: C · O(n)

Why this is the answer

A heap gives no rule for which subtree contains a key, so in the worst case all nodes must be checked; only the minimum is fast to find.

Answer

C · O(n)
5.Graph theory

Answer: Euler path, no circuit

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.

Degrees

deg(A) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 2, deg(E) = 3

Odd vertices

C, E → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

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: 12, C: 7
2. C (7): E: 19
3. B (12): D: 25
4. E (19)
5. D (25): F: 35
6. F (35)

round 4

E

Answer

E

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