CS106B Midterm Review

Transcription

1 CS106B Midterm Review Anna Zeng and Aleksander Dash Slides adapted from Anton Apostolatos, Ashley Taylor and Nolan Handali Special thanks to Brahm Capoor for everything recursion-related

2 Today s Session Overview Logistics C++ basics ADTs Big-O Recursion Tips, tricks, and everything in between! Memory, pointers, and link nodes Additional study tips

3 Logistics

4 Midterm Logistics Thursday February 15th, 7:00-9:30 pm Memorial Auditorium Open book No electronic copies, no print-outs There will be a few copies in the front if you don t have a copy Closed notes (reference sheet will be provided) Pencils highly recommended, pens accepted Problems graded based on functionality, not style But if we can t read your handwriting, we can t grade you!

5 What s on the midterm Topics: C++ basics: Console I/O, strings, streams, functions and pass by value/reference ADTs: Vector, Grid, Stack, Queue, Set (+ HashSet), Map (+ HashMap) Big-Oh Notation Recursion Backtracking Pointers 8-10 questions total Mix between reading code and writing code Source: XKCD

6 What s not on the midterm

7 C++ Basics

8 Understand what Tip #0: happens when you pass by value vs. reference Make sample mystery Tip #1: code and test yourself and your friends Review assignments 1 and 2 for I/O, strings Tip #2: and file streams Source: International Obfuscated C Code Contest

9 Tracing Exercise int main() { int frodo = 7; int sam = 5; int merry = 4; int pippin = cervantes(sam, dumas(frodo, sam, merry)); cout << sam << endl; cout << pippin << endl; cout << merry << endl; return 0; int cervantes(int &sancho, int quixote) { sancho *= quixote; return quixote--; int dumas(int athos, int &aramis, int &porthos) { if (athos + aramis < porthos) { athos = aramis - porthos; else { porthos--; return aramis - (athos + 7);

10 Tracing Exercise int main() { int frodo = 7; int sam = 5; int merry = 4; int pippin = cervantes(sam, dumas(frodo, sam, merry)); cout << sam << endl; cout << pippin << endl; cout << merry << endl; return 0; Console int cervantes(int &sancho, int quixote) { sancho *= quixote; return quixote--; int dumas(int athos, int &aramis, int &porthos) { if (athos + aramis < porthos) { athos = aramis - porthos; else { porthos--; return aramis - (athos + 7);

11 ADTs

12 43 Vector Grid (1D list of elements) (2D list of elements)

13 Stack Queue (LIFO linear structure) (FIFO linear structure) push pop enqueue dequeue

14 Map Set (Collection of key/value pairs) (Collection of unique elements)

15 ADT Exercise I Write a function that reads a input file of phone numbers and prints all the phone numbers with the most commonly occurring area code. Sample input.txt: Sample output:

16 Solution (part 1) void areacodes(string filename) { // open file ifstream input; ifstream input(filename); input.open(filename.c_str()); if (input.fail()) { return; // read file data into map of sets Map <string, Set<string>> numbers; string line; while (getline(input, line)) { string areacode = line.substr(0, 3); numbers[areacode].add(line);

17 Solution (part 2) // find most popular area code string best = ""; for (string areacode : numbers) { if (best.empty() numbers[areacode].size() > numbers[best].size()) { best = areacode; // print all numbers in that area code for (string number : numbers[best]) { cout << number << endl;

18 ADT Exercise II Write a function commoncontacts(map<string, int>& mybook, Map<string, int>& theirbook) that returns a set of the names of all the common contacts (same name & number) between two phonebooks Chris Marty Mehran

19 ADT Exercise II - Using vectors Set<string> commoncontacts(map<string, int>& mybook, Map<string, int>& theirbook) { Set<string> commoncontacts; for (string mycontact : mybook.keys()) { for (string theircontact : theirbook.keys()) { if (mycontact == theircontact && mybook[mycontact] == theirbook[theircontact]) { commoncontacts.add(mycontact); return commoncontacts;

20 Big-Oh Notation

21 Big-Oh Exercise I If vec has N elements and database has M elements: int mystery(vector<int>& vec, Set<int>& database) { int total = 0; Set<int> s = database; for (int itemone : vec) { for (int itemtwo : database) { if (itemone == itemtwo) { total++; else { for (int itemthree : vec) { s.add(itemthree); return total;

22 Big-Oh Exercise I - Answer: O(N²Mlog(M)) If vec has N elements and database has M elements: int mystery(vector<int>& vec, Set<int>& database) { int total = 0; // O(1) Set<int> s = database; // O(M) for (int itemone : vec) { // O(N * M for (int itemtwo : database) { // O(M if (itemone == itemtwo) { // total++; // else { for (int itemthree : vec) { // s.add(itemthree); return total; // O(1) * N * log(m)) * N * log(m)) O(1) O(1) O(N * log(m)) // O(log(M))

23 Big-Oh Exercise II If s has N elements: int recurse(stack<int>& s) { if (!s.isempty()){ int x = s.pop(); return x + recurse(s); return 0;

24 Big-Oh Exercise II - Answer: O(N) If s has N elements: int recurse(stack<int>& s) { if (!s.isempty()){ int x = s.pop(); return x + recurse(s); return 0; // O(1) // O(1) What is the Big-Oh of each recursive call? How many recursive calls do we make? Each recursive step is O(1), and since we go through the entire stack one element at a time, we have N recursive steps, so we multiply them to get O(1 * N)

25 Recursion Brahm Capoor In order to understand recursion, you must first understand recursion

26 Section problems 1. Problem 3: Reverse 2. Problem 5: Subsequence

27 Reverse Let s find the base case When is the problem so simple you don t need to think? What s the simplest possible string? Now let s do the harder cooler part If we have a more complex string, how can we make the problem simpler? reverse( banter ) = retnab key insight: we can break a string into two parts: the first letter, and the rest of the string How does this suggest recursion?

28 The Pseudocode function reverse(s): if s is blank: return a blank string else: c = first character of s r = rest of string return reverse(r) + c The Code string reverse(string s) { if (s == ) return ; return reverse(s.substr(1)) + s[0];

29 Subsequence: is s2 a subsequence of s1? Let s find the base case(s) When is the problem so simple that you don t need to think? What s the simplest possible string? How many base cases do we need? Now let s do the harder cooler part If we have more complex strings, how do we make the problem simpler? examples: issubsequence( catch, cat ) = issubsequence( atch, at ) issubsequence( scatter, cat ) = issubsequence( catter, cat ) key insights: if the two strings have the same first character, we examine the rest of the strings if the two strings have different first characters, we examine the rest of the first string

30 The Pseudocode function is_subseq(s1, s2) : if s2 is blank: return true if s1 is blank: return false else: if same first character: r1 = rest of s1 r2 = rest of s2 return is_subseq(r1, r2) else: r1 = rest of s1 return is_subseq(r1, s2) The Code bool is_subseq(string s1, string s2) { if (s2 == ) return true; if (s1 == ) return false; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return is_subseq(r1, r2); else { string r1 = s1.substr(1); return is_subseq(r1, s2);

31 Other useful recursion problems Tracing problem: Basic recursion practice: Problem 1 Problem 2: Sum of squares Recursion string problems: Problem 3: Reverse Problem 4: Star string

32 Backtracking Choose. Explore. Unchoose. Repeat.

33 Recursion vs backtracking: building an intuition string reverse(string s) { if (s == ) return ; return reverse(s.substr(1)) + s[0]; bool is_subseq(string s1, string s2) { if (s2 == ) return true; if (s1 == ) return false; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return is_subseq(r1, r2); else { string r1 = s1.substr(1); return is_subseq(r1, s2);

34 Recursion vs backtracking: building an intuition string reverse(string s) { if (s == ) return ; return reverse(s.substr(1)) + s[0]; bool is_subseq(string s1, string s2) { if (s2 == ) return true; if (s1 == ) return false; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return is_subseq(r1, r2); else { string r1 = s1.substr(1); return is_subseq(r1, s2); // the recursive cases are mutually exclusive In recursion, you only ever do one recursive call at every level of the recursion In recursion, you know that your recursive call will work (it s the leap of faith!)

35 Section problems 1. Problem 4: Longest Common Subsequence

36 issubsequence bool is_subseq(string s1, string s2) { if (s2 == ) return true; if (s1 == ) return false; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return is_subseq(r1, r2); else { string r1 = s1.substr(1); return is_subseq(r1, s2); subseq( banter, bar ) = subseq( anter, ar ) subseq( banter, ant ) = subseq( anter, ant ) Let s find a base case When is the problem so simple you don t need to think? What s the simplest possible string? Let s think about more complicated versions of the problem Let s try and mirror the structure of bool is_subseq(string s1, string s2) What s the LCS when the two strings have the same first character? What s the LCS when the two strings have different first characters?

37 LCS redux: so why backtracking? bool is_subseq(string s1, string s2) { if (s2 == ) return true; if (s1 == ) return false; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return is_subseq(r1, r2); else { string r1 = s1.substr(1); return is_subseq(r1, s2); subseq( banter, bar ) = subseq( anter, ar ) subseq( banter, ant ) = subseq( anter, ant ) Let s find a base case When is the problem so simple you don t need to think? What s the simplest possible string? when strings are blank! Let s think about more complicated versions of the problem Let s try and mirror the structure of bool is_subseq(string s1, string s2) What s the LCS when the two strings have the same first character? LCS(the rest) What s the LCS when the two strings have different first characters?

38 LCS redux: so why backtracking? What do we do when the two strings have different first characters? LCS( banter, ant ) = LCS( anter, ant ) How did I know to remove the b from banter? LCS( banter, abat ) = LCS( banter, bat ) How did I know to remove the a from abat? Radical idea: What if I didn t know? What if I tried both ways? I d get two subsequences. How would I know which was better? Hint: What is the function trying to find? Let s put it all together...

39 The Pseudocode Compare: issubsequence Pseudocode function LCS(s1, s2) : if s1 or s2 is blank: return a blank string else: if same first character: c = first char r1 = rest of s1 r2 = rest of s2 return c + LCS(r1, r2) else: r1 = rest of s1 r2 = rest of s2 p1 = LCS(s1, r2) p2 = LCS(r1, s2) return longer of p1 and p2 function is_subseq(s1, s2) : if s2 is blank: return true if s1 is blank: return false else: if same first character: r1 = rest of s1 r2 = rest of s2 return is_subseq(r1, r2) else: r1 = rest of s1 return is_subseq(r1, s2)

40 The Pseudocode function LCS(s1, s2) { if s1 or s2 is blank: return a blank string else: if same first character: c = first char r1 = rest of s1 r2 = rest of s2 return c + LCS(r1, r2) else: r1 = rest of s1 r2 = rest of s2 p1 = LCS(s1, r2) p2 = LCS(r1, s2) return longer of p1 and p2 The Code string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ string r1 = s1.substr(1); return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length() { return p1; else { return p2;

41 Rethinking backtracking It s hard to think of this in terms of choose/explore/unchoose But what did we do? string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

42 Rethinking backtracking It s hard to think of this in terms of choose/explore/unchoose But what did we do? We found the possible options string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

43 Rethinking backtracking It s hard to think of this in terms of choose/explore/unchoose But what did we do? We found the possible options We figured out what the best one was string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

44 Rethinking backtracking It s hard to think of this in terms of choose/explore/unchoose But what did we do? We found the possible options We figured out what the best one was We gave that one back string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

45 Rethinking backtracking It s hard to think of this in terms of choose/explore/unchoose But what did we do? We found the possible options We figured out what the best one was We gave that one back Backtracking isn t about choosing, exploring and unchoosing It s about figuring out which option works best (sometimes, that means choosing, exploring and unchoosing) string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

46 That said... Choose (kinda) string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

47 That said... Choose (kinda) Explore (kinda) string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

48 That said... Choose (kinda) Explore (kinda) Unchoose (kinda-er) string LCS(string s1, string s2) { if (s1 == s2 == ) return ; if (s1[0] == s2[0]){ return s1[0] + LCS(r1, r2); else { string r1 = s1.substr(1); string p1 = LCS(s1, r2); string p2 = LCS(r1, s2); if (p1.length() > p2.length()) { return p1; else { return p2;

49 Other useful backtracking problems (all of them) Working with the choose-explore-unchoose paradigm: Using Recursive helper functions: Problem 6: Make Change Problem 9: Ways to climb Batsh!t awesome problems: Problem 7: Print Squares Problem 10: Twiddle Problem 11: Hacking & cracking (So literally, all of them. These are awesome midterm practice)

50 endl; /* good luck with the midterm! */

51 Pointers and Linked Nodes

52 Linked Lists 101 int main() { ListNode* front = new ListNode(); front->data = 3; delete front; Use new to create a new object on the heap To access data on the other end of a pointer, use -> Remember that this is just the same as, (*front).data Need to delete whatever you create Otherwise, memory leaks!

53 struct ListNode { int data; ListNode* next; Linked Nodes practice list1 list (8 out of 9 practice midterms have some form of this as the linked list problem) Tips: Draw a picture! Potentially declare temporary pointers to help you rearrange nodes

54 Linked Nodes practice answer ListNode* temp = list1; list1 temp list2 ListNode* temp2 = list2; temp2 list1 list1 = list1->next; temp 4 list2 = list1->next; temp2 5 list2

55 Linked Nodes practice answer list1 list1 -> next = temp2 -> next; list1->next->next = temp; list1->next->next->next = nullptr; temp temp2 1 4 list2 ->next = temp2; list2->next->next = nullptr; list

56 Pointer trace tips: The * symbol has two uses type* ptrname declares a pointer that points to an object of type type e.g. ListNode* front is a pointer that points to a ListNode *ptrname dereferences ptrname (aka, goes to where the pointer is pointing to) We ve seen this! Recall that ptr->data is equivalent to (*ptr).data

57 Pointer trace tips: The & symbol has two uses &value means give me the memory address that value is stored at e.g. &value returns something that looks like 0x20065ab0 Don t confuse this with reference parameters in function prototypes! e.g. void printlist (Vector<int>& v) {...

58 Pointer trace practice (from Practice Midterm #9) What is the output?

59 Pointer trace practice (from Practice Midterm #9) What is the output? Console

60 Additional Study Tips Good practice materials: Practice exams (9 total) Leftover section problems Codestepbystep.com See Julie Zelenski s Exam Strategies handout (bottom of cs106b.stanford.edu/exams.shtml) You got this! Good luck!