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 recurrence T(n) = 2T(n/2) + O(n) solves to:

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

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

  1. AΘ(n²)
  2. BΘ(n log² n)
  3. CΘ(n log n)
  4. DΘ(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. A2i + 1
  2. B(i − 1) mod n
  3. Ci + 1
  4. D(i + 1) mod n
4.Data structuresHard

The keys 5, 3, 8, 1 are inserted in this order into an empty array-based binary min-heap. What is the final array?

  1. A[1, 3, 8, 5]
  2. B[3, 1, 8, 5]
  3. C[1, 5, 8, 3]
  4. D[1, 3, 5, 8]
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K5 have? Circuits that differ only in the starting vertex or the direction count as the same.

K5
H =
6.Graph theoryHard
ABCDEF

How many closed walks of length 3 start and end at B? 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

By the master theorem (a = 2, b = 2, f(n) = n), T(n) = Θ(n log n).

Answer

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

Answer: B · 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

B · Θ(n log² n)
3.Data structures

Answer: D · (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

D · (i + 1) mod n
4.Data structures

Answer: A · [1, 3, 8, 5]

Why this is the answer

After 5, 3, 8 the array is [3, 5, 8]; 1 goes to index 3, swaps with its parent 5 and then with 3, giving [1, 3, 8, 5].

Answer

A · [1, 3, 8, 5]
5.Graph theory

Answer: H = 12

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 = (5 − 1)!/2 = 24/2 = 12

Answer

H = 12
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 0 0 0 0
B: 1 0 1 1 1 0
C: 0 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

One more step

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

Answer

N = 4

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