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 multiplying two n × n matrices with the standard triple loop?

  1. AO(n³)
  2. BO(n² log n)
  3. CO(n)
  4. DO(n²)
2.Algorithms & complexityHard

What is the running time of Dijkstra's algorithm with a binary heap on a graph with V vertices and E edges?

  1. AO((V + E) log V)
  2. BO(V³)
  3. CO(V + E)
  4. DO(V × E)
3.Data structuresMedium

In a binary min-heap stored in an array (index starting at 0), the children of the node at index i are at:

  1. A2i + 1 and 2i + 2
  2. Bi + 1 and i + 2
  3. C2i and 2i + 1
  4. Di/2 and i/2 + 1
4.Data structuresHard

Inserting a key into an AVL tree makes a node's left subtree too tall, and the new key went into the left subtree of its left child (left-left case). Which fix is applied?

  1. ARecolouring the node and its child
  2. BA left rotation at the child, then a right rotation at the node
  3. CA single right rotation at the unbalanced node
  4. DA single left rotation at the unbalanced node
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
ABCDEF

How many closed walks of length 2 start and end at D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: A · O(n³)

Why this is the answer

There are n² entries and each needs n multiplications, giving n³ operations.

Answer

A · O(n³)
2.Algorithms & complexity

Answer: A · O((V + E) log V)

Why this is the answer

There are V extract-min and up to E decrease-key operations, each O(log V); a Fibonacci heap improves this to O(E + V log V).

Answer

A · O((V + E) log V)
3.Data structures

Answer: A · 2i + 1 and 2i + 2

Why this is the answer

For 0-based arrays, the left child is at 2i + 1, the right child at 2i + 2, and the parent at ⌊(i − 1)/2⌋.

Answer

A · 2i + 1 and 2i + 2
4.Data structures

Answer: C · A single right rotation at the unbalanced node

Why this is the answer

In the left-left case one right rotation lifts the left child up and restores balance; zig-zag cases need a double rotation.

Answer

C · A single right rotation at the unbalanced node
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: N = 3

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 0 1 1 0
C: 1 0 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)DD = 0·0 + 1·1 + 0·0 + 0·0 + 1·1 + 1·1 = 3

Answer

N = 3

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