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 amortised time complexity of appending to a dynamic array that doubles its capacity when full?

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

The 0/1 knapsack problem solved with dynamic programming takes O(nW) time. Why is this called pseudo-polynomial?

  1. ABecause n is always small
  2. BW is a number, so the time is exponential in the number of bits used to write W
  3. CBecause it only works for W = 0
  4. DBecause the algorithm is randomised and its running time varies with each input
3.Data structuresMedium

Which collision resolution method stores colliding keys in a list at the same bucket?

  1. ADouble hashing
  2. BQuadratic probing
  3. CLinear probing
  4. DSeparate chaining
4.Data structuresHard

A Fenwick tree (binary indexed tree) over an array of n numbers supports which operations in O(log n)?

  1. AFinding the median in O(1)
  2. BPrefix-sum queries and single-element updates
  3. CInserting elements at any position
  4. DSorting the array and binary search
5.Graph theoryMedium

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

2 × deg 4
L =
6.Graph theoryHard
ABCDEF

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

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · O(1)

Why this is the answer

Copying all elements happens only at powers of two; spread over all appends, the average cost per append is constant.

Answer

C · O(1)
2.Algorithms & complexity

Answer: B · W is a number, so the time is exponential in the number of bits used to write W

Why this is the answer

The running time is polynomial in the value of W but exponential in its length (log W bits); 0/1 knapsack is NP-complete.

Answer

B · W is a number, so the time is exponential in the number of bits used to write W
3.Data structures

Answer: D · Separate chaining

Why this is the answer

Chaining keeps a linked list (or other structure) per bucket; open addressing looks for another free slot in the table itself.

Answer

D · Separate chaining
4.Data structures

Answer: B · Prefix-sum queries and single-element updates

Why this is the answer

A Fenwick tree stores partial sums over ranges determined by the low bits of each index, so both prefix sums and point updates take O(log n).

Answer

B · Prefix-sum queries and single-element updates
5.Graph theory

Answer: L = 6

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 + 8 = 2(L + 2) − 2
L = 6

Answer

L = 6

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

Common neighbours

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

Answer

N = 1

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