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) = T(n/2) + O(1) solve to?

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

Which bound is correct for log(n!)?

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

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo make the list sorted
  2. BTo prevent the list from growing beyond a fixed maximum number of nodes
  3. CTo simplify code by removing special cases at the head or tail
  4. DTo store the number of elements
4.Data structuresHard

Which pair of traversals always determines a unique binary tree with distinct keys?

  1. APreorder and postorder
  2. BLevel-order alone
  3. CPreorder alone
  4. DPreorder and inorder
5.Graph theoryMedium
ABCDEFGEuler 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| = 7|E| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
32376101077ABCDEFG

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(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

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

Answer: B · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

B · Θ(n log n)
3.Data structures

Answer: C · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

C · To simplify code by removing special cases at the head or tail
4.Data structures

Answer: D · Preorder and inorder

Why this is the answer

Preorder gives the root, and its position in the inorder sequence splits the left and right subtrees; preorder with postorder is ambiguous when a node has one child.

Answer

D · Preorder and inorder
5.Graph theory

Answer: neither

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) = 3, deg(C) = 4, deg(D) = 2, deg(E) = 3, deg(F) = 1, deg(G) = 3

Odd vertices

B, E, F, G → 4
neither

Answer

neither
6.Graph theory

Answer: D

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: 3, C: 2
2. C (2): G: 12
3. B (3): D: 10
4. D (10): E: 16, F: 20
5. G (12)
6. E (16)
7. F (20)

round 4

D

Answer

D

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