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

You sort an array of n elements once and then answer k membership queries with binary search. What is the total cost?

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

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AComparisons take logarithmic time each
  2. BA decision tree with n possible outcomes must have depth at least log₂ n
  3. CEvery element must be compared at least once
  4. DBinary search is the only possible algorithm, and it needs log₂ n comparisons
3.Data structuresMedium

What happens when a hash table rehashes after its load factor exceeds a threshold?

  1. AThe existing keys are sorted in place
  2. BOnly the colliding keys are moved to a new overflow list at the end of the table
  3. CThe oldest keys are deleted to make room
  4. DA larger table is allocated and every key is inserted again
4.Data structuresHard

Under simple uniform hashing, what is the expected time of a search in a hash table with chaining, n keys and m buckets?

  1. AO(log n)
  2. BO(1 + α), where α = n / m
  3. CO(m)
  4. DO(n)
5.Graph theoryMedium

A tree has 4 vertices of degree 3, 3 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

4 × deg 33 × deg 5
L =
6.Graph theoryHard
5721129101339ABCDEFG

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: C · O(n log n + k log n)

Why this is the answer

Sorting costs O(n log n) and each query O(log n); this beats k linear scans (O(nk)) when k is large.

Answer

C · O(n log n + k log n)
2.Algorithms & complexity

Answer: B · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

B · A decision tree with n possible outcomes must have depth at least log₂ n
3.Data structures

Answer: D · A larger table is allocated and every key is inserted again

Why this is the answer

Since the index is computed modulo the table size, all keys must be rehashed into the new, larger table; with doubling this stays O(1) amortised.

Answer

D · A larger table is allocated and every key is inserted again
4.Data structures

Answer: B · O(1 + α), where α = n / m

Why this is the answer

Computing the hash costs O(1) and the expected chain length is α, so keeping α bounded by resizing gives O(1) expected time.

Answer

B · O(1 + α), where α = n / m
5.Graph theory

Answer: L = 15

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.

Handshake lemma for a tree

L + 27 = 2(L + 7) − 2
L = 15

Answer

L = 15

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: d(F) = 18

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 5, C: 7
B (5): G: 18
C (7): E: 8
E (8): D: 20, F: 18, G: 17
G (17)
F (18)
D (20)

Shortest path

A → C → E → F
d(F) = 18

Answer

d(F) = 18

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