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

What does the recurrence T(n) = T(n/2) + O(1) solve to?

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

Why does binary search not give O(log n) time on a singly linked list?

  1. ABinary search needs at least twice the memory
  2. BReaching the middle element takes O(n) steps, since there is no direct index access
  3. CComparisons on linked list nodes are slower
  4. DLinked lists cannot be sorted
3.Data structuresMedium

A binary tree has root 1 with left child 2 and right child 3; node 2 has children 4 (left) and 5 (right). What is the postorder traversal?

  1. A1 2 4 5 3
  2. B4 2 5 1 3
  3. C5 4 3 2 1
  4. D4 5 2 3 1
4.Data structuresHard

Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?

  1. ADeleting would increase the false positive rate to 100%
  2. BThose bits may also belong to other elements, which would cause false negatives
  3. CBits in a Bloom filter cannot be changed after creation
  4. DBloom filters store the full elements, and these cannot be located again once hashed
5.Graph theoryMedium

A tree has 3 vertices of degree 4, 4 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 44 × deg 53 × deg 2
L =
6.Graph theoryHard
13841221141ABCDEF

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

start: Aend: C
d(C) =

3Solutions

1.Algorithms & complexity

Answer: C · O(log n)

Why this is the answer

The problem halves at each step with constant work per step, so there are about log₂ n steps.

Answer

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

Answer: B · Reaching the middle element takes O(n) steps, since there is no direct index access

Why this is the answer

Binary search needs O(1) access to the middle; on a linked list walking there costs linear time, so the total is O(n).

Answer

B · Reaching the middle element takes O(n) steps, since there is no direct index access
3.Data structures

Answer: D · 4 5 2 3 1

Why this is the answer

Postorder processes 4 and 5 before 2, then 3, and finally the root 1.

Answer

D · 4 5 2 3 1
4.Data structures

Answer: B · Those bits may also belong to other elements, which would cause false negatives

Why this is the answer

Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.

Answer

B · Those bits may also belong to other elements, which would cause false negatives
5.Graph theory

Answer: L = 20

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 + 38 = 2(L + 10) − 2
L = 20

Answer

L = 20

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

6.Graph theory

Answer: d(C) = 21

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: 13
B (13): C: 21, D: 17, E: 14
E (14): D: 16, F: 18
D (16)
F (18)
C (21)

Shortest path

A → B → C
d(C) = 21

Answer

d(C) = 21

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