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

Breadth-first search on an unweighted graph finds:

  1. AThe minimum spanning tree
  2. BA topological order in all graphs
  3. CShortest paths in number of edges from the start node
  4. DShortest paths in graphs that contain negative edge weights
2.Algorithms & complexityHard

Which bound is correct for log(n!)?

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

How much memory does an adjacency list use for a graph with V vertices and E edges?

  1. AO(V + E)
  2. BO(E²)
  3. CO(V²)
  4. DO(V × E)
4.Data structuresHard

You must keep the k smallest values from a stream of n numbers using O(k) memory. Which approach works in O(n log k) time?

  1. AA max-heap of size k, replacing its root when a smaller value arrives
  2. BA sorted array of all n values
  3. CA FIFO queue of the last k values
  4. DA min-heap holding all n values, from which the k smallest are popped at the end
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
ABCDE

How many walks of length 2 go from A to E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

C · Shortest paths in number of edges from the start node
2.Algorithms & complexity

Answer: A · 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

A · Θ(n log n)
3.Data structures

Answer: A · O(V + E)

Why this is the answer

Each vertex has a list and each edge appears once (directed) or twice (undirected), giving O(V + E).

Answer

A · O(V + E)
4.Data structures

Answer: A · A max-heap of size k, replacing its root when a smaller value arrives

Why this is the answer

The max-heap's root is the largest kept value; any smaller new value replaces it in O(log k), so the heap always holds the k smallest so far.

Answer

A · A max-heap of size k, replacing its root when a smaller value arrives
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
6.Graph theory

Answer: N = 2

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)AE = 0·0 + 1·1 + 1·1 + 0·0 + 0·0 = 2

Answer

N = 2

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