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

What is the time complexity of: for i = 1 to n: for (j = i; j ≤ n; j = j + i) do O(1) work?

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

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(1), because the index is known
  2. BO(n), because later elements must shift
  3. CO(n²), because each element moves n times
  4. DO(log n), because of binary search
4.Data structuresHard

An object is stored as a key in a hash map, and then a field used by its hash function is changed. What is the likely result?

  1. AThe map throws an error at the moment the field changes
  2. BLooking up the object may fail because it sits in the bucket for its old hash
  3. CThe map automatically moves the key to its new bucket
  4. DNothing changes, because a key's hash is computed only once when the program starts
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
9914148213107ABCDEF

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: D · 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

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

Answer: B · O(n log n)

Why this is the answer

The total is n / 1 + n / 2 + ... + n / n = n × Hn, and the harmonic number Hn is Θ(log n).

Answer

B · O(n log n)
3.Data structures

Answer: B · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

B · O(n), because later elements must shift
4.Data structures

Answer: B · Looking up the object may fail because it sits in the bucket for its old hash

Why this is the answer

The key's location was chosen from its old hash; with a new hash, lookups search the wrong bucket, so keys should be immutable.

Answer

B · Looking up the object may fail because it sits in the bucket for its old hash
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: d(F) = 26

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: 9, C: 9
B (9): D: 23, E: 16
C (9)
E (16): D: 18, F: 26
D (18)
F (26)

Shortest path

A → B → E → F
d(F) = 26

Answer

d(F) = 26

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