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 amortised time complexity of appending to a dynamic array that doubles its capacity when full?

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

Karatsuba multiplication satisfies T(n) = 3T(n/2) + O(n). What does this solve to?

  1. AΘ(n²)
  2. BΘ(nlog₂ 3), about n1.585
  3. CΘ(n3)
  4. DΘ(n log n)
3.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i − 1) mod n
  2. Bi + 1
  3. C(i + 1) mod n
  4. D2i + 1
4.Data structuresHard

The union-find (disjoint set) structure with union by rank and path compression has an amortised cost per operation of:

  1. AO(n)
  2. BO(α(n)), the inverse Ackermann function, which is practically constant
  3. CO(n²)
  4. DO(log n) exactly
5.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 8
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
ABCDEF

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

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: A · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

A · O(1)
2.Algorithms & complexity

Answer: B · Θ(nlog₂ 3), about n1.585

Why this is the answer

Since log₂ 3 ≈ 1.585 > 1, the leaves dominate by the master theorem, beating the O(n²) schoolbook method.

Answer

B · Θ(nlog₂ 3), about n1.585
3.Data structures

Answer: C · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

C · (i + 1) mod n
4.Data structures

Answer: B · O(α(n)), the inverse Ackermann function, which is practically constant

Why this is the answer

With both optimisations, operations take O(α(n)) amortised time; α(n) ≤ 4 for any realistic n.

Answer

B · O(α(n)), the inverse Ackermann function, which is practically constant
5.Graph theory

Answer: Euler path, no circuit

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.

Degrees

deg(A) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 2, deg(E) = 4, deg(F) = 1

Odd vertices

C, F → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

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 1
E: 0 1 1 1 0 1
F: 0 0 0 1 1 0

One more step

(M³)DD = 0·0 + 1·0 + 1·0 + 2·0 + 1·1 + 1·1 = 2

Answer

N = 2

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