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 time complexity of: for i = 1 to n: for j = 1 to i: do O(1) work?

  1. AO(n)
  2. BO(n³)
  3. CO(n²)
  4. DO(n log n)
2.Algorithms & complexityHard

Using the master theorem, T(n) = 8T(n/2) + O(n²) solves to:

  1. AΘ(n²)
  2. BΘ(n² log n)
  3. CΘ(n³)
  4. DΘ(n log 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. A√n
  2. Blog₂ n
  3. Cn (it becomes a linked list)
  4. D1
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. ATo avoid repeated resizing when pushes and pops alternate near the boundary
  2. BBecause the memory allocator can only release memory in quarters of a block
  3. CBecause halving at one half would lose elements
  4. DTo keep the array sorted
5.Graph theoryMedium

A tree has 3 vertices of degree 3, 4 vertices of degree 2, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 34 × deg 2
L =
6.Graph theoryHard
710429181713ABCDEF

Prim’s algorithm builds a minimum spanning tree starting from vertex D. Click the edge it adds as edge number 5.

start: Dk = 5
  1. 1edge AB
  2. 2edge BC
  3. 3edge BD
  4. 4edge CE
  5. 5edge DE
  6. 6edge DF
  7. 7edge EF
  8. 8edge BE

3Solutions

1.Algorithms & complexity

Answer: C · O(n²)

Why this is the answer

The inner body runs n(n + 1) / 2 times in total, which is still quadratic.

Answer

C · O(n²)
2.Algorithms & complexity

Answer: C · Θ(n³)

Why this is the answer

log₂ 8 = 3, and n² grows slower than n³, so the leaves dominate: Θ(n³).

Answer

C · Θ(n³)
3.Data structures

Answer: C · 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

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

Answer: A · 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

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

Answer: L = 5

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 + 17 = 2(L + 7) − 2
L = 5

Answer

L = 5

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

6.Graph theory

Answer: edge EF

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.

Prim’s algorithm

1. BD (4)
2. AB (7)
3. DE (9)
4. CE (2)
5. EF (17)

edge number 5

EF

Answer

edge EF

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