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CS 61B: Data Structures (Autumn 2006)
Midterm II
Solutions
Problem 1. (8 points) A Miscellany. a. 20 microseconds. b. Because for all n > 101, 1 / ( n - 100) ≤ n. (Here, c = 1.) c. 3 6 / |
1 2 5 7 9 / | \ / \ / |
a b c d e f g h d. Θ(log n log log n ). That's read (log n ) (log log n ). Problem 2. (8 points) Binary Trees. a. 6 /
/
4 8 / \ /
2 5 7 9 /
1 3 b. Yes. In the following example, the preorder traversal changes from 5 3 9 7 to 7 3 9. 5 7 / \ /
3 9 ==delete(5)==> 3 9 / 7 c. 7 2 5 9 3 4 1 8 7 2 5 8 3 4 1 9 7 2 1 8 3 4 5 9 1 2 7 8 3 4 5 9
1 2 4 8 3 7 5 9 d. I would choose to be binary, Balanced, big, and written in C. Problem 3. (9 points) Directed Graphs. a. public void contract(int x, int y) { adj[x].union(adj[y]); adj[y] = new Set(); for (int i = 0; i < size; i++) { if (adj[i].remove(y)) { adj[i].insert(x); } } } For problems b-d, the union operation always takes O( s ) time, so the running time is dominated by the cost of performing one query and update on each of s data structures that together contain a total of e items. b. Θ( s + e ). c. Θ( s ). d. There are (at least) three answers we will accept for this question. The most likely reasonable answer we expect is Θ( s log e ). But the truly smallest correct bound that can be given is Θ(max { s , s log ( e / s )}). We'll also accept Θ( s + s log ( e / s )), but it's technically wrong because it can become negative when e < s. Mail inquiries to [email protected]
