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

Breadth-first search on an unweighted graph finds:

  1. AShortest paths in graphs that contain negative edge weights
  2. BThe minimum spanning tree
  3. CShortest paths in number of edges from the start node
  4. DA topological order in all graphs
2.Algorithms & complexityHard

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AAfter V − 1 rounds, it checks whether any edge can still be relaxed
  2. BIt runs Dijkstra's algorithm twice and compares the two distance results
  3. CIt checks whether any edge has a negative weight
  4. DIt counts the cycles with depth-first search
3.Data structuresMedium

A circular buffer (ring buffer) is commonly used to implement:

  1. AA graph traversal
  2. BA hash function
  3. CA binary search tree
  4. DA fixed-size queue
4.Data structuresHard

Which problem of linear probing does double hashing mainly reduce?

  1. APrimary clustering, where long runs of occupied slots form
  2. BThe need for a hash function
  3. CThe O(n) cost of reading a key's value
  4. DThe need to store keys in the table
5.Graph theoryMedium

A connected graph has 11 vertices and 19 edges. How many edges k must be removed to leave a spanning tree?

|V| = 11|E| = 19
k =
6.Graph theoryHard
1418746199ABCDE

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

|V| = 5|E| = 7
W =

3Solutions

1.Algorithms & complexity

Answer: C · Shortest paths in number of edges from the start node

Why this is the answer

BFS visits nodes in order of distance (edges) from the source, so it finds shortest unweighted paths in O(V + E).

Answer

C · Shortest paths in number of edges from the start node
2.Algorithms & complexity

Answer: A · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

A · After V − 1 rounds, it checks whether any edge can still be relaxed
3.Data structures

Answer: D · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

D · A fixed-size queue
4.Data structures

Answer: A · Primary clustering, where long runs of occupied slots form

Why this is the answer

Linear probing steps by 1, so occupied runs grow and merge; double hashing uses a second hash for the step size so different keys probe different sequences.

Answer

A · Primary clustering, where long runs of occupied slots form
5.Graph theory

Answer: k = 9

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

11 − 1 = 10

Edges to remove

k = 19 − 10 = 9

Answer

k = 9
6.Graph theory

Answer: W = 31

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

BD (4): take
CE (6): take
BC (7): take
BE (9): skip, closes a cycle
AB (14): take

Total weight

W = 4 + 6 + 7 + 14 = 31

Answer

W = 31

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