Department of Computer Science and Engineering. CSE 2011: Fundamentals of Data Structures Winter 2009, Section Z

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1 Department of omputer Science and Engineering SE 2011: Fundamentals of Data Structures Winter 2009, Section Z Instructor: N. Vlajic Date: pril 14, 2009 Midterm Examination Instructions: Examination time: 75 min. Print your name and S student number in the space provided below. This examination is closed book and closed notes. No calculators or other computing devices may be used. There are 6 questions. The points for each question are given in square brackets, next to the question title. The overall maximum score is 100. nswer each question in the space provided. If you need to continue an answer onto the back of a page, clearly indicate that and label the continuation with the question number. FIRST NME: LST NME: STUDENT #: Question Marks 1 / 20 2 / 20 3 / 24 4 / 10 5 / 14 6 / 12 1

2 1. True/False [20 points] ircle True or False, as appropriate. Question nswer 1. If f(n) O(g(n)), then g(n) is a lower bound on f(n). True False [2 points] 2. Function n log(n) grows faster than (log(n)) 2. True False [2 points] 3. f(n) = 5n n log(n) is Ω(n 3 ). True False [2 points] 4. lgorithms of complexity 2 n log(n) are called unfeasible. True False [2 points] 5. In Java, compile-time errors are generally preferred over run-time errors. True False [2 points] 6. utoboxing (in Java 5) occurs when a wrapper object is converted to the corresponding primitive type. True False [2 points] 7. The running time (RT) of binary search on a sorted array is: average case RT = O(log(n)), worst-case RT = O(n). True False [2 points] 8. Extensible array with doubling always provides better performance than extensible array with fixed increments with True False respect to set(int k, E e) 1 method. [2 points] 9. Three singly linked-lists of sizes log(n), n and n 2, respectively, can be merged into one singly linked-list in O(1) time, assuming we are not concerned with the True False ordering of elements in the resulting linked-list. [2 points] 10. When storing and accessing a set of items with equal access frequencies, a self-organizing list based on count-method strategy should always be the preferred choice over a selforganizing list based on move-to-front strategy. True False [2 points] 1 set(in k, E e new ) replaces the existing element e old at rank k with the new element e new, and returns e old. 2

3 2. Multiple hoice [20 points] ircle the letter next to the most appropriate answer. 2.1 [6 points] Recall that f(n) O(g(n)) means that c>0 and n 0, such that n n 0, f(n) c g(n) Which of the following choices of n 0 and c would work in the above definition to show that n 2 + 3n O(n 2 )? (a) (b) (c) (d) n n 0 0 = 1, c = 4. = 3, c = 2. (a) and (b). None of the above. 2.2 [4 points] Three programs P1, P2, and P3 have time complexities f 1 (n), f 2 (n), and f 3 (n), respectively, such that f1(n) is O(f 2 (n)), f2(n) is O(f 1 (n)), f1(n) is O(f 3 (n)), and f3(n) is NOT O(f 1 (n)). Which of the following statements must be true? (a) Program P 3 is always faster than P 1 and P 2. (b) Program P1 is faster than P 2 and P 3 for very large size inputs. (c) Program P1 has the exact same running time as program P 2. (d) Program P3 is slower than P 1 and P 2 for very large inputs. 2.3 [3 points] Which of the following (Java related) statements is NOT true? (a) Many variables can be bound to the same object at the same time. (b) variable can be bound to many objects at the same time. (c) variable can be bound to only one object at a time. (d) When not bound to an object, a variable has the value Null. 2.4 [3 points] Which of the following statements regarding singly linked lists is true? (a) Insert at tail, remove at head and remove at tail can be performed in constant time. (b) Insert at head, remove at head and remove at tail can be performed in constant time. (c) Insert at head, insert at tail and remove at tail can be performed in constant time. (d) Insert at head, insert at tail and remove at head can be performed in constant time. 3

4 2.5 [4 points] stack is being used to parse an html file. The parsing method makes the following calls to the stack: stack.push( <html> ); stack.push( <body> ); stack.push( <h1> ); stack.top(); stack.pop(); stack.push( <p> ); stack.top(); stack.push( <i> ); stack.top(); stack.pop(); t this point, what would be the value at the top of the stack? (a) (b) (c) (d) <html> <body> <p> <h1> 4

5 3. Running Times [24 points] (3.1) onsider the following code fragment that repeatedly calls the resize method. (resize method uses Java s System.arraycopy() to copy elements of one array into another larger array.) int[] = new int[5]; // is an integer array with five elements for (int i=0; i<n; i++) { if (i ==.length) =resize(,.length+1); [i] = 10; (3.1.a) [8 points] What is the running time of this code fragment as a function of n? Express your answer using the ig-o notation. (Show your work for partial credit!) For values of i = 0, 1, 2, 3, and 4, each execution of the body of the for-loop takes a constant, say c, number of steps because the condition in the if-statement evaluates to false. For values of i, greater than 4, each execution of the body of the for-loop takes time proportional to i. This is because the condition inside the for-loop evaluates to true and the call to the resize function, which resizes from size i to size (i+1), takes time proportional to i. Thus the total running time is: O(1) + ( n-1) = O(n^2) (3.1.b) [4 points] Making a minor change to the above code fragment will bring its running time down, dramatically. What is this change? What will the new running time be, as a function of n, in ig-o notation? Since n is known, we can replace the definition of by int[] = new int[n] If we do this, then no calls to the resize function will be made. Resulting running time: O(n). lternately, we could replace the call to the resize function by the following code: = resize(, 2*.length); s discussed in class, doing this ensures that we make only a small number of calls to resize and the overall running time is O(n). 5

6 (3.2) [6 points] What is the running time of the following code fragment, as a function of n. ssume n 1. Express the running time using big-o notation. You do not need to provide an exact analysis. int i = n; int p = 0; while (p*p < n) { i = i/2; p++; O(sqrt(n)) Since p is incremented until it passes the square root of n, and this increment cannot happen more than sqrt(n) times. Parameter i has no effect on the running time! (3.3) [6 points] What is the running time of the following code fragment, as a function of n. ssume n 1. Express the running time using big-o notation. You do not need to provide an exact analysis. int total, i, j; total = 1; i = n; while (i>0) do { j = n; while (j>0) do { total = total*10; j = j/2; i = i-10; return total; Parameter total does not have any effect on the running time. RT of inner while loop: O(log 2 (n)). RT of outer while loop: O(n/10) = O(n). Overall RT = RT outer_while RT inner_while O(nlog 2 (n)) 6

7 4. Stacks and Queues [12 points] (4.1) [6 points] Describe the contents of stack s after the method convert executes. That is, describe the contents in a general manner based on what is in s before the code executes. public void convert(stack<object> s) { rraylist<object> list = new rraylist<object>(); while (s.size() > 0) { list.add(s.pop()); for (Object o : list) { s.push(o); The order of the elements is reversed. Rank 2 Rank 1 Rank 0 Stack List Stack (4.2) [6 points] What happens if a queue is used instead of a stack in the code above, e.g., public void convert(queue<object> q) { rraylist<object> list = new rraylist<object>(); while (q.size() > 0) { list.add(q.dequeue()); for (Object o : list) { q.enqueue(o); The elements remain in the same order Rank 2 Rank 1 Rank 0 Queue List Queue 7

8 5. NodeList DT [12 points] Suppose that the NodeList class supports only the following methods: addfirst(e e) - inserts e at the beginning of the list remove(position<e> p) - removes p and returns the respective element first() - returns the first Position last() - returns the last Position size() - returns the size of the sequence Which of the following cannot be implemented using only the methods given above. (The use of additional methods and data structures in not allowed!) (a) invert() - a method that reverses the order of the elements in the node list (b) Stack - a class with methods: push, pop, and top (c) Queue - a class with methods: enqueue, dequeue, and first (d) Deque - a class with methods: addfirst, addlast, removefirst, removelast, first and last NodeList Stack pop() = remove(first()) peek() = first() push(p) = addfirst(p) Queue enqueue(p) = addfirst(p) dequeue() = remove(last()) first() = front() = first() insert here access/remove here access/remove here Deque first() = first() last() = last() addfirst(p) = addfirst(p) removefirst() = remove(first()) addlast() = { addfirst() for (int i=1; i<size(); i++) { addfirst(remove(last())); removelast() = remove(last()) 8

9 6. lgorithm Design [12 points] ssume a sorted array () of size n. Propose an algorithm for finding two elements x and y in that minimize x-y. Your algorithm should run in O(n) time for full credit. (Note: x-y represents the absolute value of (x-y).) 1, 3, 6, 9, 10, 15, 21, Since the array is sorted, the pair that has minimum difference (i.e. minimizes x-y ) will be in adjacent positions. Hence, we traverse from start to finish keeping track of the smallest difference a[j+1] - a[j]. int x=a[0]; int y=a[1]; int min = Math.abs(x-y) ; for (int i=1 ; i<(a.size()-2) ; i++) { if (min>math.abs(a[i]-a[i+1]) { min = Math.abs(a[i]-a[i+1]); x=a[i]; y=a[i+1]; return x, y 9