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 does the recurrence T(n) = 2T(n/2) + O(1) solve to?

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

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AIt checks whether any edge has a negative weight
  2. BIt runs Dijkstra's algorithm twice and compares the two distance results
  3. CAfter V − 1 rounds, it checks whether any edge can still be relaxed
  4. DIt counts the cycles with depth-first search
3.Data structuresMedium

A circular buffer (ring buffer) is commonly used to implement:

  1. AA graph traversal
  2. BA hash function
  3. CA binary search tree
  4. DA fixed-size queue
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. AStore the front and rear in the same variable
  2. BAlways clear the buffer whenever front equals rear, so the state is empty
  3. CKeep a separate element count, or always leave one slot unused
  4. DMake the buffer size a power of two
5.Graph theoryMedium

A tree has 2 vertices of degree 3, 2 vertices of degree 4, 2 vertices of degree 5, 3 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

2 × deg 32 × deg 42 × deg 53 × deg 2
L =
6.Graph theoryHard
ABCDEF

How many walks of length 2 go from A to D? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: D · O(n)

Why this is the answer

By the master theorem nlog₂ 2 = n dominates the constant work, so T(n) = Θ(n), as when finding a maximum by splitting in halves.

Answer

D · O(n)
2.Algorithms & complexity

Answer: C · 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

C · After V − 1 rounds, it checks whether any edge can still be relaxed
3.Data structures

Answer: D · A fixed-size queue

Why this is the answer

A ring buffer uses head and tail indices that wrap around, giving O(1) enqueue and dequeue in fixed memory.

Answer

D · A fixed-size queue
4.Data structures

Answer: C · 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

C · Keep a separate element count, or always leave one slot unused
5.Graph theory

Answer: L = 14

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.

Handshake lemma for a tree

L + 30 = 2(L + 9) − 2
L = 14

Answer

L = 14

Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.

6.Graph theory

Answer: N = 1

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 0
E: 0 1 1 1 0 1
F: 0 0 0 0 1 0

Common neighbours

(M²)AD = 0·0 + 1·1 + 1·0 + 0·0 + 0·1 + 0·0 = 1

Answer

N = 1

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