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 is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?

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

Which statement about little-o notation is correct?

  1. An² = o(n)
  2. B2n = o(n), because 2n grows slowly
  3. Cn = o(n), since every function is little-o of itself
  4. Dn = o(n²), but n is not o(n)
3.Data structuresMedium

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

  1. AO(V + E)
  2. BO(log V)
  3. CO(V²)
  4. DO(E)
4.Data structuresHard

You must keep the k smallest values from a stream of n numbers using O(k) memory. Which approach works in O(n log k) time?

  1. AA max-heap of size k, replacing its root when a smaller value arrives
  2. BA sorted array of all n values
  3. CA FIFO queue of the last k values
  4. DA min-heap holding all n values, from which the k smallest are popped at the end
5.Graph theoryMedium

A connected graph has 14 vertices and 23 edges. How many edges k must be removed to leave a spanning tree?

|V| = 14|E| = 23
k =
6.Graph theoryHard
ABCDEF

How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.

M = adjacency matrixL = 3
N =

3Solutions

1.Algorithms & complexity

Answer: B · O(mn)

Why this is the answer

The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.

Answer

B · O(mn)
2.Algorithms & complexity

Answer: D · 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

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

Answer: C · 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

C · O(V²)
4.Data structures

Answer: A · A max-heap of size k, replacing its root when a smaller value arrives

Why this is the answer

The max-heap's root is the largest kept value; any smaller new value replaces it in O(log k), so the heap always holds the k smallest so far.

Answer

A · A max-heap of size k, replacing its root when a smaller value arrives
5.Graph theory

Answer: k = 10

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

14 − 1 = 13

Edges to remove

k = 23 − 13 = 10

Answer

k = 10
6.Graph theory

Answer: N = 4

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

One more step

(M³)EE = 1·0 + 1·1 + 1·0 + 2·1 + 3·0 + 1·1 = 4

Answer

N = 4

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