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

A recurrence T(n) = 2T(n/2) + O(n) solves to:

  1. AO(n)
  2. BO(log n)
  3. CO(n log n)
  4. DO(n²)
2.Algorithms & complexityHard

What did the Cook–Levin theorem establish?

  1. AEvery NP problem can be solved in exponential time
  2. BThe Boolean satisfiability problem (SAT) is NP-complete
  3. CThe halting problem is undecidable
  4. DP is not equal to NP
3.Data structuresMedium

You have a reference to a node in the middle of a list. In which list can you remove it in O(1)?

  1. AA singly linked list
  2. BA sorted array
  3. CA doubly linked list
  4. DA dynamic array
4.Data structuresHard

A segment tree is typically used to:

  1. ASort strings alphabetically by storing each prefix in a separate tree node
  2. BImplement a stack
  3. CAnswer range queries (such as range sums) and updates in O(log n)
  4. DStore a graph's edges
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
8561641519ABCDE

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

|V| = 5|E| = 7
W =

3Solutions

1.Algorithms & complexity

Answer: C · O(n log n)

Why this is the answer

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

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

Answer: B · The Boolean satisfiability problem (SAT) is NP-complete

Why this is the answer

Cook and Levin showed independently that every problem in NP reduces to SAT in polynomial time, so SAT was the first known NP-complete problem.

Answer

B · The Boolean satisfiability problem (SAT) is NP-complete
3.Data structures

Answer: C · A doubly linked list

Why this is the answer

A doubly linked node knows its predecessor, so you can relink prev and next directly; a singly linked list must search for the predecessor.

Answer

C · A doubly linked list
4.Data structures

Answer: C · Answer range queries (such as range sums) and updates in O(log n)

Why this is the answer

A segment tree stores aggregates for intervals, supporting range queries and point updates in O(log n) time.

Answer

C · Answer range queries (such as range sums) and updates in O(log n)
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: W = 30

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

CE (4): take
AC (5): take
BC (6): take
AB (8): skip, closes a cycle
DE (15): take

Total weight

W = 4 + 5 + 6 + 15 = 30

Answer

W = 30

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