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 does the recurrence T(n) = T(n − 1) + O(1) solve to?

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

A problem is NP-complete if it is:

  1. AOutside NP but harder than all NP problems, such as the halting problem
  2. BSolvable in polynomial time
  3. CIn NP and every NP problem reduces to it in polynomial time
  4. DImpossible to solve
3.Data structuresMedium

An adjacency matrix for a graph with V vertices uses how much memory?

  1. AO(V²)
  2. BO(log V)
  3. CO(V + E)
  4. DO(E)
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. ABecause the memory allocator can only release memory in quarters of a block
  2. BTo keep the array sorted
  3. CTo avoid repeated resizing when pushes and pops alternate near the boundary
  4. DBecause halving at one half would lose elements
5.Graph theoryMedium

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

2 × deg 4
L =
6.Graph theoryHard
1417111631472ABCDEF

Find the total weight W of a minimum spanning tree of the weighted graph.

|V| = 6|E| = 9
W =

3Solutions

1.Algorithms & complexity

Answer: B · O(n)

Why this is the answer

There are n levels with constant work each, as in a recursive sum of an array.

Answer

B · O(n)
2.Algorithms & complexity

Answer: C · In NP and every NP problem reduces to it in polynomial time

Why this is the answer

NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.

Answer

C · In NP and every NP problem reduces to it in polynomial time
3.Data structures

Answer: A · O(V²)

Why this is the answer

An adjacency matrix stores V × V entries, efficient for dense graphs; adjacency lists use O(V + E), better for sparse graphs.

Answer

A · O(V²)
4.Data structures

Answer: C · 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

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

Answer: L = 6

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 + 8 = 2(L + 2) − 2
L = 6

Answer

L = 6

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

6.Graph theory

Answer: W = 24

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

DE (1): take
BE (2): take
CE (3): take
DF (4): take
EF (7): skip, closes a cycle
BC (11): skip, closes a cycle
AB (14): take

Total weight

W = 1 + 2 + 3 + 4 + 14 = 24

Answer

W = 24

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