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

Which statement about little-o notation is correct?

  1. An² = o(n)
  2. Bn = o(n²), but n is not o(n)
  3. C2n = o(n), because 2n grows slowly
  4. Dn = o(n), since every function is little-o of itself
3.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(n), because later elements must shift
  2. BO(log n), because of binary search
  3. CO(n²), because each element moves n times
  4. DO(1), because the index is known
4.Data structuresHard

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

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

A connected graph has 16 vertices and 27 edges. How many edges k must be removed to leave a spanning tree?

|V| = 16|E| = 27
k =
6.Graph theoryHard
7212152214122ABCDEFG

Use Dijkstra’s algorithm to find the length d(G) of the shortest path from A to G in the weighted graph.

start: Aend: G
d(G) =

3Solutions

1.Algorithms & complexity

Answer: A · 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

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

Answer: B · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

B · n = o(n²), but n is not o(n)
3.Data structures

Answer: A · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

A · O(n), because later elements must shift
4.Data structures

Answer: C · 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

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

Answer: k = 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.

Edges of a spanning tree

16 − 1 = 15

Edges to remove

k = 27 − 15 = 12

Answer

k = 12
6.Graph theory

Answer: d(G) = 6

Shortest paths and spanning trees: Dijkstra’s algorithm settles one vertex per round: the unsettled vertex with the smallest tentative distance. Its distance is then final, and every neighbour v gets d(v) = min(d(v), d(u) + w(u, v)). Kruskal’s algorithm builds a minimum spanning tree by taking the edges from light to heavy and skipping any edge that would close a cycle; Prim’s algorithm grows one tree from a start vertex by always adding the lightest edge that leaves it. A spanning tree of n vertices has n − 1 edges.

Dijkstra round by round

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 7, C: 2
C (2): E: 4
E (4): F: 18, G: 6
G (6): D: 18
B (7)
D (18)
F (18)

Shortest path

A → C → E → G
d(G) = 6

Answer

d(G) = 6

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