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. BNo subproblems at all, so it can be solved in one step
  3. CA single loop
  4. DOverlapping subproblems and optimal substructure
2.Algorithms & complexityHard

How does the Bellman–Ford algorithm detect a negative-weight cycle reachable from the source?

  1. AIt runs Dijkstra's algorithm twice and compares the two distance results
  2. BAfter V − 1 rounds, it checks whether any edge can still be relaxed
  3. CIt counts the cycles with depth-first search
  4. DIt checks whether any edge has a negative weight
3.Data structuresMedium

What is the cost of inserting an element in the middle of an array of n elements?

  1. AO(log n), because of binary search
  2. BO(n), because later elements must shift
  3. CO(n²), because each element moves n times
  4. DO(1), because the index is known
4.Data structuresHard

An AVL node is unbalanced because its left child's right subtree became too tall (left-right case). Which fix is applied?

  1. AA single right rotation at the node
  2. BA single left rotation at the node
  3. CA left rotation at the left child, then a right rotation at the node
  4. DA right rotation at the left child, then a left rotation at the node
5.Graph theoryMedium
ABCDEFEuler circuitEuler path, no circuitneither

Does the connected graph have an Euler circuit, an Euler path that is not a circuit, or neither? Click your answer.

|V| = 6|E| = 9
  1. 1Euler circuit
  2. 2Euler path, no circuit
  3. 3neither
6.Graph theoryHard
6896969555ABCDEFG

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: D · 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

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

Answer: B · After V − 1 rounds, it checks whether any edge can still be relaxed

Why this is the answer

All true shortest distances are found within V − 1 rounds; if an edge can still be improved, a negative cycle must exist.

Answer

B · After V − 1 rounds, it checks whether any edge can still be relaxed
3.Data structures

Answer: B · O(n), because later elements must shift

Why this is the answer

All elements after the insertion point move one position to the right, which takes time proportional to n.

Answer

B · O(n), because later elements must shift
4.Data structures

Answer: C · A left rotation at the left child, then a right rotation at the node

Why this is the answer

The first rotation converts the left-right case into a left-left case, which a right rotation at the unbalanced node then fixes.

Answer

C · A left rotation at the left child, then a right rotation at the node
5.Graph theory

Answer: Euler path, no circuit

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.

Degrees

deg(A) = 2, deg(B) = 4, deg(C) = 3, deg(D) = 3, deg(E) = 4, deg(F) = 2

Odd vertices

C, D → 2
Euler path, no circuit

Answer

Euler path, no circuit

The Euler path starts at one odd vertex and ends at the other.

6.Graph theory

Answer: d(F) = 21

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: 6
B (6): C: 14, D: 15, G: 11
G (11): E: 16
C (14)
D (15): F: 21
E (16)
F (21)

Shortest path

A → B → D → F
d(F) = 21

Answer

d(F) = 21

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