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

  1. AO(n²)
  2. BO(n)
  3. CO(n³)
  4. DO(n log n)
2.Algorithms & complexityHard

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n³)
  2. BΘ(n²)
  3. CΘ(n² log n)
  4. DΘ(n log n)
3.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. ALevel order
  2. BInsertion order
  3. CDescending order always
  4. DAscending sorted order
4.Data structuresHard

How does a radix tree (compressed trie) improve on a plain trie?

  1. AIt stores each word in a hash table at the root instead of in separate nodes
  2. BIt merges chains of single-child nodes into one edge labelled with a string
  3. CIt sorts the children of each node by frequency
  4. DIt keeps only the first letter of each word
5.Graph theoryMedium

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

4 × deg 42 × deg 5
L =
6.Graph theoryHard
ABCDEF

How many closed walks of length 3 start and end at B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: A · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

A · O(n²)
2.Algorithms & complexity

Answer: A · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

A · Θ(n³)
3.Data structures

Answer: D · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

D · Ascending sorted order
4.Data structures

Answer: B · It merges chains of single-child nodes into one edge labelled with a string

Why this is the answer

Collapsing paths with no branching reduces the number of nodes and the memory used while keeping prefix search.

Answer

B · It merges chains of single-child nodes into one edge labelled with a string
5.Graph theory

Answer: L = 16

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 + 26 = 2(L + 6) − 2
L = 16

Answer

L = 16

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

6.Graph theory

Answer: N = 4

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 0 0 0 0
B: 1 0 1 1 1 0
C: 0 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)BB = 0·1 + 4·0 + 1·1 + 1·1 + 2·1 + 2·0 = 4

Answer

N = 4

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