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

Dynamic programming is useful when a problem has:

  1. AOverlapping subproblems and optimal substructure
  2. BNo subproblems at all, so it can be solved in one step
  3. CA single loop
  4. DOnly random inputs
2.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n³)
  2. BO(n² 2ⁿ)
  3. CO(2n²)
  4. DO(n!)
3.Data structuresMedium

What is the worst-case height of a binary search tree with n nodes if keys are inserted in sorted order?

  1. An (it becomes a linked list)
  2. B1
  3. C√n
  4. Dlog₂ n
4.Data structuresHard

You must keep the k smallest values from a stream of n numbers using O(k) memory. Which approach works in O(n log k) time?

  1. AA max-heap of size k, replacing its root when a smaller value arrives
  2. BA min-heap holding all n values, from which the k smallest are popped at the end
  3. CA sorted array of all n values
  4. DA FIFO queue of the last k values
5.Graph theoryMedium

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

2 × deg 52 × deg 2
L =
6.Graph theoryHard
15146128111312148ABCDEFG

Dijkstra’s algorithm runs from A. Click the vertex that is settled (made permanent) in round 6; A itself is round 1.

start: Around 6
  1. 1A
  2. 2B
  3. 3C
  4. 4D
  5. 5E
  6. 6F
  7. 7G

3Solutions

1.Algorithms & complexity

Answer: A · Overlapping subproblems and optimal substructure

Why this is the answer

Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.

Answer

A · Overlapping subproblems and optimal substructure
2.Algorithms & complexity

Answer: B · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

B · O(n² 2ⁿ)
3.Data structures

Answer: A · n (it becomes a linked list)

Why this is the answer

Sorted insertions create a degenerate tree with height n, making searches O(n). Balanced trees prevent this.

Answer

A · n (it becomes a linked list)
4.Data structures

Answer: A · A max-heap of size k, replacing its root when a smaller value arrives

Why this is the answer

The max-heap's root is the largest kept value; any smaller new value replaces it in O(log k), so the heap always holds the k smallest so far.

Answer

A · A max-heap of size k, replacing its root when a smaller value arrives
5.Graph theory

Answer: L = 8

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 + 14 = 2(L + 4) − 2
L = 8

Answer

L = 8

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

6.Graph theory

Answer: G

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: 1, C: 5
2. B (1): D: 7
3. C (5): E: 17, G: 17
4. D (7): E: 15, F: 18
5. E (15)
6. G (17)
7. F (18)

round 6

G

Answer

G

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