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

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. ANo, because n³ is smaller than n² for large n
  2. BYes, but only if n is smaller than 100
  3. CYes, because Big O is an upper bound and n² grows no faster than n³
  4. DNo, because Big O must always state the exact growth rate of the function
2.Algorithms & complexityHard

Which algorithm finds a minimum spanning tree by repeatedly adding the cheapest edge that does not form a cycle?

  1. AKruskal's algorithm
  2. BPrim's algorithm starting from one node
  3. CDijkstra's algorithm
  4. DFloyd–Warshall
3.Data structuresMedium

What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?

  1. A√n
  2. Blog₂ n
  3. C1
  4. Dn (it becomes a linked list)
4.Data structuresHard

What guarantee do the red-black tree rules give about the height of a tree with n keys?

  1. AThe height is at most n / 2
  2. BThe height is at most log2 n + 1
  3. CThe height is exactly log2 n
  4. DThe height is at most 2 log2(n + 1)
5.Graph theoryMedium

A connected graph has 15 vertices and 22 edges. How many edges k must be removed to leave a spanning tree?

|V| = 15|E| = 22
k =
6.Graph theoryHard
ABCDEF

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

C · Yes, because Big O is an upper bound and n² grows no faster than n³
2.Algorithms & complexity

Answer: A · Kruskal's algorithm

Why this is the answer

Kruskal sorts edges by weight and adds them if they connect different components; Prim grows one tree from a start vertex.

Answer

A · Kruskal's algorithm
3.Data structures

Answer: D · n (it becomes a linked list)

Why this is the answer

Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.

Answer

D · n (it becomes a linked list)
4.Data structures

Answer: D · The height is at most 2 log2(n + 1)

Why this is the answer

Every path has the same number of black nodes and red nodes cannot be adjacent, so the longest path is at most twice the shortest, giving height O(log n).

Answer

D · The height is at most 2 log2(n + 1)
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

15 − 1 = 14

Edges to remove

k = 22 − 14 = 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 1 0 0
C: 1 1 0 0 0 0
D: 0 1 0 0 1 1
E: 0 0 0 1 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)AA = 0·0 + 1·1 + 1·1 + 0·0 + 0·0 + 0·0 = 2

Answer

N = 2

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