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

Is it correct to say that an O(n²) algorithm is also O(n³)?

  1. AYes, because Big O is an upper bound and n² grows no faster than n³
  2. BNo, because n³ is smaller than n² for large n
  3. CNo, because Big O must always state the exact growth rate of the function
  4. DYes, but only if n is smaller than 100
2.Algorithms & complexityHard

Which technique gives an expected O(n log n) running time for quicksort on any input?

  1. ASorting the input before running quicksort
  2. BUsing recursion only on the left side
  3. CAlways choosing the first element
  4. DChoosing the pivot uniformly at random
3.Data structuresMedium

How long does it take to check whether the edge (u, v) exists using an adjacency matrix?

  1. AO(E)
  2. BO(log V)
  3. CO(V)
  4. DO(1)
4.Data structuresHard

In a circular buffer, front == rear can mean either empty or full. What is a common way to tell them apart?

  1. AKeep a separate element count, or always leave one slot unused
  2. BStore the front and rear in the same variable
  3. CMake the buffer size a power of two
  4. DAlways clear the buffer whenever front equals rear, so the state is empty
5.Graph theoryMedium

A connected graph has 20 vertices and 28 edges. How many edges k must be removed to leave a spanning tree?

|V| = 20|E| = 28
k =
6.Graph theoryHard
11081116132ABCDE

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

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

3Solutions

1.Algorithms & complexity

Answer: A · Yes, because Big O is an upper bound and n² grows no faster than n³

Why this is the answer

Big O is an upper bound, so the statement is true but not tight; Θ(n²) states the exact growth rate.

Answer

A · Yes, because Big O is an upper bound and n² grows no faster than n³
2.Algorithms & complexity

Answer: D · Choosing the pivot uniformly at random

Why this is the answer

With a random pivot, the expected running time is O(n log n) for every input; bad cases still exist but are very unlikely.

Answer

D · Choosing the pivot uniformly at random
3.Data structures

Answer: D · O(1)

Why this is the answer

The matrix entry at row u, column v tells you immediately whether the edge exists; an adjacency list may need to scan u's neighbours.

Answer

D · O(1)
4.Data structures

Answer: A · Keep a separate element count, or always leave one slot unused

Why this is the answer

With a count, empty means count = 0 and full means count = capacity; with one empty slot, full is defined as (rear + 1) mod n == front.

Answer

A · Keep a separate element count, or always leave one slot unused
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

20 − 1 = 19

Edges to remove

k = 28 − 19 = 9

Answer

k = 9
6.Graph theory

Answer: W = 22

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

AB (1): take
BE (2): take
BC (8): take
AC (10): skip, closes a cycle
BD (11): take

Total weight

W = 1 + 2 + 8 + 11 = 22

Answer

W = 22

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