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) = T(n − 1) + O(1) solve to?

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

What does the recurrence T(n) = 2T(n/2) + n log n solve to?

  1. AΘ(n²)
  2. BΘ(n log n)
  3. CΘ(n log² n)
  4. DΘ(n)
3.Data structuresMedium

A hash table of size 7 uses h(k) = k mod 7 and linear probing. After inserting 10, where is 17 stored?

  1. AIndex 3
  2. BIndex 0
  3. CIndex 4
  4. DIndex 6
4.Data structuresHard

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

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

A connected graph has 9 vertices and 10 edges. How many edges k must be removed to leave a spanning tree?

|V| = 9|E| = 10
k =
6.Graph theoryHard
ABCDEF

How many closed walks of length 2 start and end at B? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

3Solutions

1.Algorithms & complexity

Answer: C · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

C · O(n)
2.Algorithms & complexity

Answer: C · Θ(n log² n)

Why this is the answer

Level k does about n log(n / 2ᵏ) work, and summing over log n levels gives Θ(n log² n), the extended second case of the master theorem.

Answer

C · Θ(n log² n)
3.Data structures

Answer: C · Index 4

Why this is the answer

10 mod 7 = 3 and 17 mod 7 = 3; index 3 is taken, so linear probing tries the next slot, index 4.

Answer

C · Index 4
4.Data structures

Answer: A · 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

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

Answer: k = 2

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

9 − 1 = 8

Edges to remove

k = 10 − 8 = 2

Answer

k = 2
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 0
B: 1 0 1 1 1 0
C: 1 1 0 0 1 0
D: 0 1 0 0 1 1
E: 0 1 1 1 0 0
F: 0 0 0 1 0 0

Common neighbours

(M²)BB = 1·1 + 0·0 + 1·1 + 1·1 + 1·1 + 0·0 = 4

Answer

N = 4

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