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

In a sorted array, you look for two elements that sum to a target by moving one pointer from each end inwards. What is the time complexity?

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

Insertion sort runs in O(n) on sorted input. Does this contradict the Ω(n log n) lower bound for comparison sorting?

  1. AYes, the lower bound is therefore wrong
  2. BNo, the lower bound applies to the worst case over all inputs, not to every input
  3. CNo, because insertion sort is not a comparison sort
  4. DYes, but only for arrays shorter than 100
3.Data structuresMedium

In a binary min-heap stored in an array (index starting at 0), the children of the node at index i are at:

  1. A2i and 2i + 1
  2. Bi + 1 and i + 2
  3. C2i + 1 and 2i + 2
  4. Di/2 and i/2 + 1
4.Data structuresHard

In a hash table using open addressing, why is deletion often done with a "tombstone" marker?

  1. ADeleted keys must be kept forever for security, so slots can never be reused
  2. BTombstones speed up hashing
  3. CIt makes the table sorted
  4. DEmptying the slot could break the probe sequence for other keys
5.Graph theoryMedium

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

3 × deg 32 × deg 2
L =
6.Graph theoryHard
12361816204ABCDEF

Kruskal’s algorithm builds a minimum spanning tree. Click the edge it adds as edge number 3; skipped edges do not count.

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

3Solutions

1.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

If the sum is too small the left pointer moves right, if too large the right pointer moves left, so there are at most n steps.

Answer

B · O(n)
2.Algorithms & complexity

Answer: B · No, the lower bound applies to the worst case over all inputs, not to every input

Why this is the answer

The bound says every comparison sort needs about n log n comparisons on some input of size n; special inputs can be much faster.

Answer

B · No, the lower bound applies to the worst case over all inputs, not to every input
3.Data structures

Answer: C · 2i + 1 and 2i + 2

Why this is the answer

For 0-based arrays, the left child is at 2i + 1, the right child at 2i + 2, and the parent at ⌊(i − 1)/2⌋.

Answer

C · 2i + 1 and 2i + 2
4.Data structures

Answer: D · Emptying the slot could break the probe sequence for other keys

Why this is the answer

If a deleted slot became empty, later searches might stop early and miss keys that were placed further along the probe chain.

Answer

D · Emptying the slot could break the probe sequence for other keys
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 + 13 = 2(L + 5) − 2
L = 5

Answer

L = 5

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

6.Graph theory

Answer: edge BE

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. BD (1)
2. AC (3)
3. BE (4)
4. BC (6)
5. DF (16)

edge number 3

BE

Answer

edge BE

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