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

Why does binary search not give O(log n) time on a singly linked list?

  1. AReaching the middle element takes O(n) steps, since there is no direct index access
  2. BBinary search needs at least twice the memory
  3. CComparisons on linked list nodes are slower
  4. DLinked lists cannot be sorted
3.Data structuresMedium

In a circular queue stored in an array of capacity n, how is the next position after index i computed?

  1. Ai + 1
  2. B2i + 1
  3. C(i + 1) mod n
  4. D(i − 1) mod n
4.Data structuresHard

The time complexity of decrease-key in a binary heap (given the element's position) is:

  1. AO(n)
  2. BO(n log n)
  3. CO(1)
  4. DO(log n)
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
ABCDEF

How many closed walks of length 2 start and end at F? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 2
N =

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 · Reaching the middle element takes O(n) steps, since there is no direct index access

Why this is the answer

Binary search needs O(1) access to the middle; on a linked list walking there costs linear time, so the total is O(n).

Answer

A · Reaching the middle element takes O(n) steps, since there is no direct index access
3.Data structures

Answer: C · (i + 1) mod n

Why this is the answer

The modulo operation wraps index n − 1 back to 0, so the array is reused without shifting elements.

Answer

C · (i + 1) mod n
4.Data structures

Answer: D · O(log n)

Why this is the answer

After decreasing the key, the element sifts up at most the height of the heap. Fibonacci heaps achieve O(1) amortised decrease-key.

Answer

D · O(log n)
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: N = 2

Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.

Adjacency matrix

A: 0 1 0 0 0 0
B: 1 0 1 1 1 0
C: 0 1 0 0 0 0
D: 0 1 0 0 1 1
E: 0 1 0 1 0 1
F: 0 0 0 1 1 0

Common neighbours

(M²)FF = 0·0 + 0·0 + 0·0 + 1·1 + 1·1 + 0·0 = 2

Answer

N = 2

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