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

Shortest simple paths can be found in polynomial time. What is known about finding the longest simple path between two vertices in a general graph?

  1. AIt is NP-hard
  2. BIt is undecidable
  3. CIt can be solved by Dijkstra's algorithm with negated weights
  4. DIt is always solvable in O(V + E)
3.Data structuresMedium

Why can a binary heap be stored in an array without any pointers?

  1. AArrays automatically keep the keys in sorted order, so no pointers are needed
  2. BHeaps never contain more than 100 elements
  3. CAll keys in a heap are equal
  4. DIt is a complete binary tree, so positions follow simple index formulas
4.Data structuresHard

An AVL node is unbalanced because its left child's right subtree became too tall (left-right case). Which fix is applied?

  1. AA right rotation at the left child, then a left rotation at the node
  2. BA single left rotation at the node
  3. CA single right rotation at the node
  4. DA left rotation at the left child, then a right rotation at the 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
ABCDE

How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: B · 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

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

Answer: A · It is NP-hard

Why this is the answer

A longest simple path of length V − 1 is a Hamiltonian path, an NP-complete problem; on directed acyclic graphs, however, it is solvable in linear time.

Answer

A · It is NP-hard
3.Data structures

Answer: D · It is a complete binary tree, so positions follow simple index formulas

Why this is the answer

Completeness means there are no gaps, so the children of index i are always at 2i + 1 and 2i + 2.

Answer

D · It is a complete binary tree, so positions follow simple index formulas
4.Data structures

Answer: D · A left rotation at the left child, then a right rotation at the node

Why this is the answer

The first rotation converts the left-right case into a left-left case, which a right rotation at the unbalanced node then fixes.

Answer

D · A left rotation at the left child, then a right rotation at the 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 = 4

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
B: 1 0 1 1 1
C: 1 1 0 0 1
D: 0 1 0 0 1
E: 0 1 1 1 0

One more step

(M³)EE = 2·0 + 2·1 + 1·1 + 1·1 + 3·0 = 4

Answer

N = 4

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