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 does it mean that a problem has optimal substructure?

  1. AThe problem can be solved in constant time
  2. BAn optimal solution can be built from optimal solutions of its subproblems
  3. CThe problem has exactly one optimal solution, which every subproblem shares
  4. DEvery subproblem has the same size
2.Algorithms & complexityHard

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

  1. AΘ(n log² n)
  2. BΘ(n log n)
  3. CΘ(n²)
  4. DΘ(n)
3.Data structuresMedium

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo make the list sorted
  2. BTo prevent the list from growing beyond a fixed maximum number of nodes
  3. CTo store the number of elements
  4. DTo simplify code by removing special cases at the head or tail
4.Data structuresHard

Why can elements not simply be deleted from a standard Bloom filter by clearing their bits?

  1. AThose bits may also belong to other elements, which would cause false negatives
  2. BBloom filters store the full elements, and these cannot be located again once hashed
  3. CBits in a Bloom filter cannot be changed after creation
  4. DDeleting would increase the false positive rate to 100%
5.Graph theoryMedium

A connected graph has 18 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?

|V| = 18|E| = 18
k =
6.Graph theoryHard
ABCDEF

How many walks of length 2 go from E to F? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: B · An optimal solution can be built from optimal solutions of its subproblems

Why this is the answer

Optimal substructure is required for both greedy algorithms and dynamic programming to produce correct answers.

Answer

B · An optimal solution can be built from optimal solutions of its subproblems
2.Algorithms & complexity

Answer: A · 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

A · Θ(n log² n)
3.Data structures

Answer: D · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

D · To simplify code by removing special cases at the head or tail
4.Data structures

Answer: A · Those bits may also belong to other elements, which would cause false negatives

Why this is the answer

Clearing a shared bit would make another element look absent; counting Bloom filters keep counters instead of bits to allow deletion.

Answer

A · Those bits may also belong to other elements, which would cause false negatives
5.Graph theory

Answer: k = 1

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

18 − 1 = 17

Edges to remove

k = 18 − 17 = 1

Answer

k = 1
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 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 0
F: 0 0 0 1 0 0

Common neighbours

(M²)EF = 0·0 + 1·0 + 1·0 + 1·1 + 0·0 + 0·0 = 1

Answer

N = 1

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