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 the Bellman–Ford algorithm on a graph with V vertices and E edges?

  1. AO(V × E)
  2. BO(V + E)
  3. CO(V³ × E)
  4. DO(E log V)
2.Algorithms & complexityHard

Which bound is correct for log(n!)?

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

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

  1. ATo simplify code by removing special cases at the head or tail
  2. BTo prevent the list from growing beyond a fixed maximum number of nodes
  3. CTo make the list sorted
  4. DTo store the number of elements
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. CBits in a Bloom filter cannot be changed after creation
  4. DThose bits may also belong to other elements, which would cause false negatives
5.Graph theoryMedium

A connected graph has 18 vertices and 23 edges. How many edges k must be removed to leave a spanning tree?

|V| = 18|E| = 23
k =
6.Graph theoryHard
111969815164ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 8
W =

3Solutions

1.Algorithms & complexity

Answer: A · O(V × E)

Why this is the answer

Each of the V − 1 rounds relaxes all E edges, giving O(VE); in exchange it handles negative edge weights.

Answer

A · O(V × E)
2.Algorithms & complexity

Answer: B · Θ(n log n)

Why this is the answer

log(n!) ≤ n log n, and the largest n / 2 factors give log(n!) ≥ (n / 2) log(n / 2); this is why comparison sorting needs Ω(n log n).

Answer

B · Θ(n log n)
3.Data structures

Answer: A · 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

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

Answer: D · 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

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

Answer: k = 6

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 = 23 − 17 = 6

Answer

k = 6
6.Graph theory

Answer: W = 45

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.

Kruskal’s algorithm

BE (4): take
BD (6): take
DE (8): skip, closes a cycle
CE (9): take
AC (11): take
DF (15): take

Total weight

W = 4 + 6 + 9 + 11 + 15 = 45

Answer

W = 45

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