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

What is the time complexity of the Bellman–Ford algorithm on a graph with V vertices and E edges?

  1. AO(V³ × E)
  2. BO(V × E)
  3. CO(E log V)
  4. DO(V + E)
2.Algorithms & complexityHard

Which statement about selecting the k-th smallest element of an unsorted array is correct?

  1. AQuickselect runs in O(log n) worst case, because it recurses on only one side of the pivot
  2. BSelection always needs Ω(n log n) because it requires sorting
  3. CQuickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
  4. DSelection is only possible in O(n²)
3.Data structuresMedium

A circular buffer (ring buffer) is commonly used to implement:

  1. AA fixed-size queue
  2. BA graph traversal
  3. CA hash function
  4. DA binary search tree
4.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n²)
  2. BO(α(n)), the inverse Ackermann function, which is practically constant
  3. CO(log n) exactly
  4. DO(n)
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
6.Graph theoryHard
ABCDE

How many closed walks of length 2 start and end at C? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: B · O(V × E)

Why this is the answer

Each of the V − 1 rounds relaxes all E edges, giving O(VE); in exchange it handles negative edge weights.

Answer

B · O(V × E)
2.Algorithms & complexity

Answer: C · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case

Why this is the answer

Quickselect's expected work is n + n/2 + n/4 + ... = O(n) with random pivots; choosing the pivot as a median of medians makes the worst case linear.

Answer

C · Quickselect runs in expected O(n), and median-of-medians guarantees O(n) worst case
3.Data structures

Answer: A · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

A · A fixed-size queue
4.Data structures

Answer: B · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

B · O(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theory

Answer: H = 3

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 = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
6.Graph theory

Answer: N = 3

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 1 0 0
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 0
E: 0 1 1 0 0

Common neighbours

(M²)CC = 1·1 + 1·1 + 0·0 + 0·0 + 1·1 = 3

Answer

N = 3

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