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

Dynamic programming is useful when a problem has:

  1. AOnly random inputs
  2. BOverlapping subproblems and optimal substructure
  3. CNo subproblems at all, so it can be solved in one step
  4. DA single loop
2.Algorithms & complexityHard

The Held–Karp dynamic programming algorithm solves the travelling salesman problem exactly. What is its running time for n cities?

  1. AO(n² 2ⁿ)
  2. BO(n!)
  3. CO(n³)
  4. DO(2n²)
3.Data structuresMedium

What is the purpose of a sentinel (dummy) node in a linked list?

  1. ATo prevent the list from growing beyond a fixed maximum number of nodes
  2. BTo simplify code by removing special cases at the head or tail
  3. CTo store the number of elements
  4. DTo make the list sorted
4.Data structuresHard

In the BST with root 50, left subtree 30 (children 20, 40) and right subtree 70 (children 60, 80), the root is deleted using the in-order successor. Which key becomes the new root?

  1. A80
  2. B70
  3. C60
  4. D40
5.Graph theoryMedium

A connected graph has 10 vertices and 11 edges. How many edges k must be removed to leave a spanning tree?

|V| = 10|E| = 11
k =
6.Graph theoryHard
13411911169ABCDE

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

|V| = 5|E| = 7
W =

3Solutions

1.Algorithms & complexity

Answer: B · Overlapping subproblems and optimal substructure

Why this is the answer

Storing results of overlapping subproblems (memoisation or tabulation) avoids repeated work, turning exponential algorithms into polynomial ones.

Answer

B · Overlapping subproblems and optimal substructure
2.Algorithms & complexity

Answer: A · O(n² 2ⁿ)

Why this is the answer

There are about n 2ⁿ states, each computed in O(n); this is exponential but far faster than trying all (n − 1)! tours.

Answer

A · O(n² 2ⁿ)
3.Data structures

Answer: B · To simplify code by removing special cases at the head or tail

Why this is the answer

With a sentinel, every real node has a predecessor, so inserting or deleting at the head needs no special handling.

Answer

B · To simplify code by removing special cases at the head or tail
4.Data structures

Answer: C · 60

Why this is the answer

The leftmost node of the right subtree is 60, so 60 replaces 50 and its old position is removed; using the predecessor would give 40.

Answer

C · 60
5.Graph theory

Answer: k = 2

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.

Edges of a spanning tree

10 − 1 = 9

Edges to remove

k = 11 − 9 = 2

Answer

k = 2
6.Graph theory

Answer: W = 30

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

BC (1): take
AC (4): take
BE (9): take
CE (11): skip, closes a cycle
AB (13): skip, closes a cycle
DE (16): take

Total weight

W = 1 + 4 + 9 + 16 = 30

Answer

W = 30

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