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 − 1) + O(1) solve to?

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

What is the running time of Dijkstra's algorithm with a binary heap on a graph with V vertices and E edges?

  1. AO((V + E) log V)
  2. BO(V + E)
  3. CO(V × E)
  4. DO(V³)
3.Data structuresMedium

How is a new element inserted into a binary heap?

  1. ARebuild the whole heap from scratch
  2. BPlace it at the root and sift it down
  3. CInsert it at its sorted position by shifting all of the larger elements one place right
  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 the memory allocator can only release memory in quarters of a block
  2. BTo keep the array sorted
  3. CTo avoid repeated resizing when pushes and pops alternate near the boundary
  4. DBecause halving at one half would lose elements
5.Graph theoryMedium

A connected graph has 9 vertices and 18 edges. How many edges k must be removed to leave a spanning tree?

|V| = 9|E| = 18
k =
6.Graph theoryHard
21597127611ABCDEF

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 3; A itself is round 1.

start: Around 3
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F

3Solutions

1.Algorithms & complexity

Answer: B · O(2ⁿ)

Why this is the answer

The number of calls doubles with each level and there are n levels, so the total is about 2ⁿ, as in the Towers of Hanoi.

Answer

B · O(2ⁿ)
2.Algorithms & complexity

Answer: A · O((V + E) log V)

Why this is the answer

There are V extract-min and up to E decrease-key operations, each O(log V); a Fibonacci heap improves this to O(E + V log V).

Answer

A · O((V + E) log V)
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: C · 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

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

Answer: k = 10

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 = 18 − 8 = 10

Answer

k = 10
6.Graph theory

Answer: D

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.

Dijkstra round by round

1. A (0): B: 2, C: 15
2. B (2): C: 11, D: 9
3. D (9): E: 16, F: 15
4. C (11)
5. F (15)
6. E (16)

round 3

D

Answer

D

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