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(n³)
  3. CO(n log n)
  4. DO(n)
2.Algorithms & complexityHard

What is the amortised cost of incrementing a binary counter, given that one increment may flip many bits?

  1. AO(log n) per increment
  2. BO(n log n) per increment
  3. CO(1) per increment
  4. DO(n) per increment
3.Data structuresMedium

A node in a binary search tree has a right subtree. Where is its in-order successor?

  1. AIts right child, always
  2. BThe rightmost node of its left subtree
  3. CIts parent, always
  4. DThe leftmost node of its right subtree
4.Data structuresHard

Which property must a red-black tree satisfy?

  1. AEvery node must be red
  2. BEvery path from a node to its descendant leaves has the same number of black nodes
  3. CEvery red node must have red children, so red nodes always form unbroken chains to the leaves
  4. DThe root must be red
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K6 have? Circuits that differ only in the starting vertex or the direction count as the same.

K6
H =
6.Graph theoryHard
ABCDEF

How many closed walks of length 2 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
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: C · O(1) per increment

Why this is the answer

Over n increments, bit i flips about n / 2ⁱ times, so the total is less than 2n flips, an average of O(1) each.

Answer

C · O(1) per increment
3.Data structures

Answer: D · The leftmost node of its right subtree

Why this is the answer

All larger keys close to the node are in its right subtree, and the smallest of those is the leftmost node there.

Answer

D · The leftmost node of its right subtree
4.Data structures

Answer: B · Every path from a node to its descendant leaves has the same number of black nodes

Why this is the answer

Red-black rules: nodes are red or black, root is black, red nodes have black children, and all paths have equal black-height, keeping height ≤ 2 log₂(n + 1).

Answer

B · Every path from a node to its descendant leaves has the same number of black nodes
5.Graph theory

Answer: H = 60

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 = (6 − 1)!/2 = 120/2 = 60

Answer

H = 60
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 1 0 0 0
B: 1 0 1 1 1 0
C: 1 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

Common neighbours

(M²)EE = 0·0 + 1·1 + 1·1 + 1·1 + 0·0 + 1·1 = 4

Answer

N = 4

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