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. BA topological order in all graphs
  3. CShortest paths in number of edges from the start node
  4. DThe minimum spanning tree
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 counts the cycles with depth-first search
  4. DIt checks whether any edge has a negative weight
3.Data structuresMedium

An in-order traversal of a binary search tree visits the keys in:

  1. ALevel order
  2. BAscending sorted order
  3. CInsertion order
  4. DDescending order always
4.Data structuresHard

In the BST with root 50, left subtree 30 (children 20, 40) and right subtree 70 (children 60, 80), the root is deleted using the in-order successor. Which key becomes the new root?

  1. A80
  2. B40
  3. C60
  4. D70
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K4 have? Circuits that differ only in the starting vertex or the direction count as the same.

K4
H =
6.Graph theoryHard
221413811121331414ABCDEFG

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

start: Aend: F
d(F) =

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: B · Ascending sorted order

Why this is the answer

Visiting the left subtree, then the node, then the right subtree outputs BST keys in sorted order.

Answer

B · Ascending sorted order
4.Data structures

Answer: C · 60

Why this is the answer

The leftmost node of the right subtree is 60, so 60 replaces 50 and its old position is removed; using the predecessor would give 40.

Answer

C · 60
5.Graph theory

Answer: H = 3

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.

Count

H = (4 − 1)!/2 = 6/2 = 3

Answer

H = 3
6.Graph theory

Answer: d(F) = 23

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: 2, C: 2
B (2): D: 15, G: 5
C (2): E: 10
G (5)
E (10): F: 23
D (15)
F (23)

Shortest path

A → C → E → F
d(F) = 23

Answer

d(F) = 23

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