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

Quicksort always uses the first element as pivot. On which input does it show its worst-case behaviour?

  1. AAn array whose elements are all different
  2. BAn array that is already sorted
  3. CAn array in random order
  4. DAn array of length 2
2.Algorithms & complexityHard

Why does any comparison-based search in a sorted array need Ω(log n) comparisons in the worst case?

  1. AComparisons take logarithmic time each
  2. BBinary search is the only possible algorithm, and it needs log₂ n comparisons
  3. CEvery element must be compared at least once
  4. DA decision tree with n possible outcomes must have depth at least log₂ n
3.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. A(i + 1) mod n
  2. B(i − 1) mod n
  3. C2i + 1
  4. Di + 1
4.Data structuresHard

Which pair of traversals always determines a unique binary tree with distinct keys?

  1. APreorder and inorder
  2. BPreorder alone
  3. CPreorder and postorder
  4. DLevel-order alone
5.Graph theoryMedium

How many different Hamilton circuits H does the complete graph K7 have? Circuits that differ only in the starting vertex or the direction count as the same.

K7
H =
6.Graph theoryHard
512731719162ABCDEF

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 5; skipped edges do not count.

k = 5
  1. 1edge AB
  2. 2edge AC
  3. 3edge BC
  4. 4edge BD
  5. 5edge CE
  6. 6edge DE
  7. 7edge EF
  8. 8edge BE

3Solutions

1.Algorithms & complexity

Answer: B · An array that is already sorted

Why this is the answer

On sorted input the pivot is always the minimum, so one side is empty and the other has n − 1 elements, giving O(n²).

Answer

B · An array that is already sorted
2.Algorithms & complexity

Answer: D · A decision tree with n possible outcomes must have depth at least log₂ n

Why this is the answer

Each comparison at most halves the set of remaining possibilities, so distinguishing n positions needs at least log₂ n comparisons.

Answer

D · A decision tree with n possible outcomes must have depth at least log₂ n
3.Data structures

Answer: A · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

A · (i + 1) mod n
4.Data structures

Answer: A · Preorder and inorder

Why this is the answer

Preorder gives the root, and its position in the inorder sequence splits the left and right subtrees; preorder with postorder is ambiguous when a node has one child.

Answer

A · Preorder and inorder
5.Graph theory

Answer: H = 360

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.

Count

H = (7 − 1)!/2 = 720/2 = 360

Answer

H = 360
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.

Kruskal’s algorithm

1. BE (2)
2. BD (3)
3. AB (5)
4. BC (7)
5. EF (16)

edge number 5

EF

Answer

edge EF

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