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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

Which statement about a recursive function without a reachable base case is correct?

  1. AIt runs in O(1) time
  2. BIt keeps calling itself until the call stack overflows
  3. CThe compiler always turns it into a loop that ends safely
  4. DIt returns 0 automatically
2.Algorithms & complexityHard

How do n! and 2ⁿ compare for large n?

  1. AThey grow at the same rate
  2. B2ⁿ grows faster than n!
  3. Cn! grows faster than 2ⁿ
  4. Dn! is O(n²)
3.Data structuresMedium

A queue is implemented with two stacks: push onto an "in" stack, and pop from an "out" stack, refilling it from "in" when empty. What is the amortised cost per operation?

  1. AO(n)
  2. BO(n²)
  3. CO(1)
  4. DO(log n)
4.Data structuresHard

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. AA left rotation at the child, then a right rotation at the node
  2. BRecolouring the node and its child
  3. CA single left rotation at the unbalanced node
  4. DA single right rotation at the unbalanced node
5.Graph theoryMedium

A connected graph has 15 vertices and 23 edges. How many edges k must be removed to leave a spanning tree?

|V| = 15|E| = 23
k =
6.Graph theoryHard
131816121014172011573ABCDEFG

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 7|E| = 12
W =

3Solutions

1.Algorithms & complexity

Answer: B · It keeps calling itself until the call stack overflows

Why this is the answer

Without a base case the recursion never stops; in practice memory for stack frames runs out and the program crashes.

Answer

B · It keeps calling itself until the call stack overflows
2.Algorithms & complexity

Answer: C · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

C · n! grows faster than 2ⁿ
3.Data structures

Answer: C · O(1)

Why this is the answer

Each element is pushed and popped at most twice in total, so although one refill can take O(n), the average over a sequence is O(1).

Answer

C · O(1)
4.Data structures

Answer: D · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

D · A single right rotation at the unbalanced node
5.Graph theory

Answer: k = 9

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

15 − 1 = 14

Edges to remove

k = 23 − 14 = 9

Answer

k = 9
6.Graph theory

Answer: W = 51

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.

Kruskal’s algorithm

GB (1): take
GE (3): take
GD (7): take
CE (10): take
BD (12): skip, closes a cycle
AB (13): take
DE (14): skip, closes a cycle
GC (15): skip, closes a cycle
BC (16): skip, closes a cycle
DF (17): take

Total weight

W = 1 + 3 + 7 + 10 + 13 + 17 = 51

Answer

W = 51

ExamProblems · examproblems.com/courses/data-structures-and-algorithms