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

How do n! and 2ⁿ compare for large n?

  1. A2ⁿ grows faster than n!
  2. Bn! is O(n²)
  3. Cn! grows faster than 2ⁿ
  4. DThey grow at the same rate
3.Data structuresMedium

Which data structure is used by depth-first search (iterative version)?

  1. AA heap
  2. BA hash table only
  3. CA queue
  4. DA stack
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[3, 1, 8, 5]
  2. B[1, 3, 8, 5]
  3. C[1, 3, 5, 8]
  4. D[1, 5, 8, 3]
5.Graph theoryMedium
ABCDEEuler 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| = 5|E| = 5
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
ABCDEF

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

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: D · 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

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

Answer: C · n! grows faster than 2ⁿ

Why this is the answer

For n ≥ 4, n! exceeds 2ⁿ, and the ratio grows without bound because each extra factor of n! is larger than 2.

Answer

C · n! grows faster than 2ⁿ
3.Data structures

Answer: D · A stack

Why this is the answer

DFS explores as deep as possible, naturally using a stack (or recursion's call stack); BFS uses a queue.

Answer

D · A stack
4.Data structures

Answer: B · [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

B · [1, 3, 8, 5]
5.Graph theory

Answer: Euler 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) = 2, deg(C) = 2, deg(D) = 2, deg(E) = 2

Odd vertices

0
Euler circuit

Answer

Euler circuit
6.Graph theory

Answer: N = 1

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 0 1 0 0 0
B: 0 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

One more step

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

Answer

N = 1

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