Name (First and Last): SAMPLE SOLUTIONS

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1 Name (First and Last): SAMPLE SOLUTIONS CS0: Data structures & algorithms Midterm Exam March 1, 017 You may use double-sided sheets of 8.5x11 paper for notes. No electronics. Show your work when appropriate. We highly recommended that you use pencil not pen. Incomplete answers are better than leaving a problem blank. Wait for the signal to start the exam before you open the exam. Take a deep breath and relax. Question Points 0 / 1 1 / / 10 / 8 TOTAL / 5 Question 0 (1 point) a) If the following statement is true, then write "I did not give or receive help in the taking of this exam and will not discuss the exam until /" and sign your name. Hello world b) draw a mark on the line to indicate how you feel before the exam c) Test # of student to your left (or aisle): Cat Test # of student to your right (or aisle): Dog Do not open the test before given permission

2 1. ( points) What is the running time? Give the running time of each function in Θ notation. You may use the following page for work if needed. // give answer in terms of n public static void f1(int n) { for (int i = 0; i < 10*n; i += 1) { System.out.println("CS0"); System.out.println("Rocks"); // give answer in terms of m and n // where m = length of the array public static void f(int n, int[] arr) { for (int i = 0; i < n; i += 1) { for (int j = arr.length-1; j >= 0; j -= 1) { arr[j] = arr[j] + 1; // give answer in terms of n // where n = length of the array f is called on public static int f(int[] arr) { if (arr.length == 0) return 0; if (arr.length == 1) return arr[0]; Answer Θ(n) Θ(n m) Θ(nlogn) int[] left = new int[arr.length/+1]; int[] right = new int[arr.length/+1]; copy(arr, left, 0, arr.length/); copy(arr, right, arr.length/, arr.length); return f(left) - f(right); // helper method is Θ x where x is length of "to" private static void copy(int[] from, int[] to, int start, int end) { int j = 0; for (int i=start; i<end; i++) { to[j] = from[i]; j++;

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4 . (10 points) Processing lists Suppose we want to build an iterator PresentInList that produces each element from a List input that is present in an unsorted array wanted. Assume that input does not contain any null elements. Example: input: 9,,, 7, 7, 1 wanted:, 7, 5 PresentInList produces, 7, 7 Consider the unfinished implementation of PresentInList below. class PresentInList<T> implements Iterator<T> { private final Iterator<T> inputiter; private final T[] wanted; private T nextelement; public PresentInList(List<T> input, T[] wanted) { this.inputiter = input.iterator(); this.wanted = wanted; this.nextelement = null; findnext(); /* fixes the "interesting invariant" */ private void findnext() {... public boolean hasnext() { return nextelement!= null; public T next() { if (!hasnext()) throw new NoSuchElementException(); T result = nextelement; nextelement = null; findnext(); return result; a) ( points) Given the partial implementation above, what is the "interesting invariant" that hasnext() and next() are taking advantage of? nextelement!= null!fromiter.hasnext()

5 Name (First and Last): b) ( points) Fill in the implementation of findnext. It will help to consider the invariant. private void findnext() { while (nextelement == null && inputiter.hasnext()) { T temp = inputiter.next(); for (int i=0; i<arr.length; i++) { if (arr[i].equals(temp)) { nextelement = temp; break; c) ( points) What is the worst case running time of one call to findnext()? n = length of input m = length of wanted O(n m) d) ( points) What is the worst case running time of the following code? n = length of the List X m = length of wanted PresentInList<Integer> myiter = new PresentInList<>(X); while(myiter.hasnext()) { System.out.println(myIter.next()); O(n m)

6 . (8 points) Tree matching. (8 points) Tree matching Type equation here. class TreeNode { int data; TreeNode left; TreeNode right; Consider the problem of finding an instance of a subtree within a larger binary tree rooted at r. Here is the method signature for TreeNode.findPattern() /* returns true if the tree rooted at pattern exists somewhere within the tree rooted at r */ static boolean findpattern(treenode r, TreeNode pattern) We wrote test cases. Note that the test cases are small, but findpattern must work for trees of any size and shape. T1) findpattern(, ) returns true T) findpattern( null, null ) returns true T) findpattern(, ) returns true T) findpattern(, ) returns true

7 Name (First and Last): T5) findpattern(, ) returns false class TreeNode { int data; TreeNode left; TreeNode right; T) findpattern(, null ) returns true a) ( points) We came up with the following code and ran the tests above in order, but it failed when it got to test case T. Modify the code just enough to pass test cases T1, T, and T. In the left margin write an A where your fix begins. b) ( points) After your fix, tests T1,T,T pass but then test T fails. Modify the code just enough so that test T also passes (T5 and T should end up passing, too). In the left margin write a B where your fix begins. static boolean findpattern(treenode r, TreeNode pattern) { if (r == null) return pattern == null; B if (pattern == null) return true; if (r.data == pattern.data && findpattern(r.left, pattern.left) && findpattern(r.right, pattern.right)) { return true; A return findpattern(r.left, pattern) findpattern(r.right, pattern); return false; // okay if you left this in