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

Why is the average lookup in a hash table O(1) but the worst case O(n)?

  1. AMany keys may collide into the same bucket
  2. BKeys must be sorted first
  3. CHash functions are slow
  4. DArrays have O(n) access
3.Data structuresMedium

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

  1. ADescending order always
  2. BInsertion order
  3. CLevel order
  4. DAscending sorted order
4.Data structuresHard

What are the time and extra space costs of reversing a singly linked list in place?

  1. AO(n) time and O(1) extra space
  2. BO(n) time and O(n) extra space
  3. CO(1) time and O(1) extra space
  4. DO(n log n) time and O(1) extra space
5.Graph theoryMedium

A connected graph has 11 vertices and 22 edges. How many edges k must be removed to leave a spanning tree?

|V| = 11|E| = 22
k =
6.Graph theoryHard
ABCDE

How many walks of length 3 go from D to 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 · Many keys may collide into the same bucket

Why this is the answer

With a good hash function and load factor, buckets stay short on average; adversarial or poor hashing can put all keys in one bucket.

Answer

A · Many keys may collide into the same bucket
3.Data structures

Answer: D · 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

D · Ascending sorted order
4.Data structures

Answer: A · O(n) time and O(1) extra space

Why this is the answer

Keeping three pointers (previous, current, next) lets you reverse each link in one pass without allocating new nodes.

Answer

A · O(n) time and O(1) extra space
5.Graph theory

Answer: k = 12

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 = 22 − 10 = 12

Answer

k = 12
6.Graph theory

Answer: N = 5

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)DE = 1·0 + 1·1 + 2·1 + 2·1 + 1·0 = 5

Answer

N = 5

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