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

A greedy algorithm:

  1. ATries every possible solution
  2. BAlways gives the optimal solution
  3. CMakes the locally best choice at each step
  4. DUses only recursion
2.Algorithms & complexityHard

Which statement about little-o notation is correct?

  1. A2n = o(n), because 2n grows slowly
  2. Bn² = o(n)
  3. Cn = o(n²), but n is not o(n)
  4. Dn = o(n), since every function is little-o of itself
3.Data structuresMedium

How does extract-min work in a binary min-heap?

  1. ASearch all leaves for the new minimum
  2. BRemove the root and shift all elements one place left
  3. CRemove the last element of the array
  4. DMove the last element to the root and sift it down
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 halving at one half would lose elements
  3. CBecause the memory allocator can only release memory in quarters of a block
  4. DTo keep the array sorted
5.Graph theoryMedium

A tree has 3 vertices of degree 4, 2 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?

3 × deg 42 × deg 5
L =
6.Graph theoryHard
91411415352ABCDEF

Use Dijkstra’s algorithm to find the length d(F) of the shortest path from A to F in the weighted graph.

start: Aend: F
d(F) =

3Solutions

1.Algorithms & complexity

Answer: C · Makes the locally best choice at each step

Why this is the answer

Greedy algorithms are fast and optimal for some problems (e.g. Dijkstra, Huffman coding) but not in general (e.g. coin change with some coin systems).

Answer

C · Makes the locally best choice at each step
2.Algorithms & complexity

Answer: C · n = o(n²), but n is not o(n)

Why this is the answer

f = o(g) means f(n) / g(n) tends to 0; n / n² tends to 0, but n / n = 1 does not.

Answer

C · n = o(n²), but n is not o(n)
3.Data structures

Answer: D · Move the last element to the root and sift it down

Why this is the answer

The last element fills the hole at the root and is swapped with its smaller child until the heap property holds, taking O(log n).

Answer

D · Move the last element to the root and sift it down
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 = 14

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 + 22 = 2(L + 5) − 2
L = 14

Answer

L = 14

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

6.Graph theory

Answer: d(F) = 16

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

Each line: the vertex settled, its final distance, and the neighbours whose distance improves.

A (0): B: 9, C: 14
B (9): D: 20, E: 11
E (11): F: 16
C (14)
F (16): D: 19
D (19)

Shortest path

A → B → E → F
d(F) = 16

Answer

d(F) = 16

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