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 greedy algorithm:

  1. AMakes the locally best choice at each step
  2. BTries every possible solution
  3. CUses only recursion
  4. DAlways gives the optimal solution
2.Algorithms & complexityHard

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

  1. AΘ(n log² n)
  2. BΘ(n log n)
  3. CΘ(n)
  4. DΘ(n²)
3.Data structuresMedium

How long does a search for a word of length L take in a trie with a fixed alphabet?

  1. AO(L × n)
  2. BO(n), where n is the number of words
  3. CO(L), independent of how many words are stored
  4. DO(log n), where n is the number of words
4.Data structuresHard

Which data structure supports an LRU (least recently used) cache with O(1) get and put?

  1. AA stack only
  2. BA single sorted array
  3. CA binary search tree only
  4. DA hash map combined with a doubly linked list
5.Graph theoryMedium

A connected graph has 20 vertices and 27 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 27
k =
6.Graph theoryHard
ABCDEF

How many walks of length 3 go from E to A? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: A · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

A · Makes the locally best choice at each step
2.Algorithms & complexity

Answer: A · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

A · Θ(n log² n)
3.Data structures

Answer: C · O(L), independent of how many words are stored

Why this is the answer

Each character selects one child, so a lookup takes L steps no matter how many words the trie contains.

Answer

C · O(L), independent of how many words are stored
4.Data structures

Answer: D · A hash map combined with a doubly linked list

Why this is the answer

The hash map finds nodes in O(1) and the doubly linked list moves accessed items to the front and evicts from the back in O(1).

Answer

D · A hash map combined with a doubly linked list
5.Graph theory

Answer: k = 8

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.

Edges of a spanning tree

20 − 1 = 19

Edges to remove

k = 27 − 19 = 8

Answer

k = 8
6.Graph theory

Answer: N = 2

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

One more step

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

Answer

N = 2

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