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: (j = 1; while j < n: j = j × 2)?

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

Which bound is correct for log(n!)?

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

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ALinear probing
  2. BDouble hashing
  3. CSeparate chaining
  4. DQuadratic probing
4.Data structuresHard

Which pair of traversals always determines a unique binary tree with distinct keys?

  1. APreorder alone
  2. BLevel-order alone
  3. CPreorder and inorder
  4. DPreorder and postorder
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
ABCDE

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

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: A · O(n log n)

Why this is the answer

The inner loop runs about log₂ n times for each of the n outer iterations, giving O(n log n).

Answer

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

Answer: A · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

A · Θ(n log n)
3.Data structures

Answer: C · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

C · Separate chaining
4.Data structures

Answer: C · Preorder and inorder

Why this is the answer

Preorder gives the root, and its position in the inorder sequence splits the left and right subtrees; preorder with postorder is ambiguous when a node has one child.

Answer

C · Preorder and inorder
5.Graph theory

Answer: H = 360

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 = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
6.Graph theory

Answer: N = 5

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

One more step

(M³)EC = 2·1 + 1·0 + 0·0 + 1·0 + 3·1 = 5

Answer

N = 5

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