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

Why is the base of a logarithm usually left out in Big O, as in O(log n)?

  1. AThe base only matters for small n
  2. BAll logarithms in computer science have base 10, so it is implied
  3. CLogarithms with different bases differ only by a constant factor
  4. DLogarithms are always rounded to integers
2.Algorithms & complexityHard

Which algorithm finds a minimum spanning tree by repeatedly adding the cheapest edge that does not form a cycle?

  1. APrim's algorithm starting from one node
  2. BFloyd–Warshall
  3. CKruskal's algorithm
  4. DDijkstra's algorithm
3.Data structuresMedium

How is a new element inserted into a binary heap?

  1. AInsert it at its sorted position by shifting all of the larger elements one place right
  2. BPlace it at the root and sift it down
  3. CRebuild the whole heap from scratch
  4. DPlace it at the end of the array and sift it up while it violates the heap property
4.Data structuresHard

A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?

  1. ABecause halving at one half would lose elements
  2. BTo avoid repeated resizing when pushes and pops alternate near the boundary
  3. CTo keep the array sorted
  4. DBecause the memory allocator can only release memory in quarters of a block
5.Graph theoryMedium

A tree has 3 vertices of degree 3, 2 vertices of degree 4, 3 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?

3 × deg 32 × deg 43 × deg 53 × deg 2
L =
6.Graph theoryHard
161598722015ABCDEFG

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

|V| = 7|E| = 9
W =

3Solutions

1.Algorithms & complexity

Answer: C · Logarithms with different bases differ only by a constant factor

Why this is the answer

Changing the base multiplies the logarithm by a constant, and Big O ignores constant factors.

Answer

C · Logarithms with different bases differ only by a constant factor
2.Algorithms & complexity

Answer: C · Kruskal's algorithm

Why this is the answer

Kruskal sorts edges by weight and adds them if they connect different components; Prim grows one tree from a start vertex.

Answer

C · Kruskal's algorithm
3.Data structures

Answer: D · Place it at the end of the array and sift it up while it violates the heap property

Why this is the answer

Adding at the end keeps the tree complete; swapping with the parent restores the order in O(log n) steps.

Answer

D · Place it at the end of the array and sift it up while it violates the heap property
4.Data structures

Answer: B · To avoid repeated resizing when pushes and pops alternate near the boundary

Why this is the answer

Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).

Answer

B · To avoid repeated resizing when pushes and pops alternate near the boundary
5.Graph theory

Answer: L = 18

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 + 38 = 2(L + 11) − 2
L = 18

Answer

L = 18

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

6.Graph theory

Answer: W = 41

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

BC (1): take
GC (2): take
BD (5): take
GB (7): skip, closes a cycle
EF (8): take
DF (9): take
GE (15): skip, closes a cycle
AC (16): take

Total weight

W = 1 + 2 + 5 + 8 + 9 + 16 = 41

Answer

W = 41

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