Section 2. ADTs and Big-O

Section problems draw upon materials from previous quarters.

This week’s section exercises explore ADT and Big-O problems. By the end of this week, you'll be well-versed in all kinds of ADTs, and you'll be able to think critically about program runtime. Runtime? More like fun-times, am I right?

Remember that the goal of section is not to get through every problem. Instead, you will cover a few problems more thoroughly, leaving the rest for you to review (and perhaps use as study material)!

Every week we will also be releasing a Qt Creator project containing starter code and testing infrastructure for that week's section problems. When a problem name is followed by the name of a .cpp file, that means you can practice writing the code for that problem in the named file of the Qt Creator project. Here is the zip of the section starter code:

📦 Starter code

1) References Available Upon Request

Topic: Reference parameters, range-based for loops

void printVector(Vector<int>& values) {
    for (int elem: values) {
        cout << elem << " ";
    }
    cout << endl;
}

void maui(Vector<int> values) {
    for (int i = 0; i < values.size(); i++) {
        values[i] = 1258 * values[i] * (values[2] - values[0]);
    }
}

void moana(Vector<int>& values) {
    for (int elem: values) {
        elem *= 137;
    }
}

void heihei(Vector<int>& values) {
    for (int& elem: values) {
        elem++;
    }
}

Vector<int> teFiti(Vector<int>& values) {
    Vector<int> result;
    for (int elem: values) {
        result += (elem * 137);
    }
    return result;
}

int main() {
    Vector<int> values = { 1, 3, 7 };
    maui(values);
    printVector(values);
    moana(values);
    printVector(values);
    heihei(values);
    printVector(values);
    teFiti(values);
    printVector(values);
    return 0;
}
Solution

Here’s the output from the program:

1 3 7
1 3 7
2 4 8
2 4 8

Here’s a breakdown of where this comes from:

  • The maui function takes its argument by value, so it’s making changes to a copy of the original vector, not the vector itself. That means that the values are unchanged back in main.
  • The moana function uses a range-based for loop to access the elements of the vector. This makes a copy of each element of the vector, so the changes made in the loop only change the temporary copy and not the elements of the vector. That makes that the values are unchanged back in main.
  • heihei, on the other hand, uses int& as its type for the range-based for loop, so in a sense it’s really iterating over the elements of the underlying vector. Therefore, its changes stick.
  • The teFiti function creates and returns a new vector with a bunch of updated values, but the return value isn’t captured back in main.

2) Debugging Deduplicating (deduplicate.cpp)

Topic: Vectors, strings, debugging

Consider the following incorrect C++ function, which accepts as input a Vector<string> and tries to modify it by removing adjacent duplicate elements:

// BUGGY!!
void deduplicate(Vector<string> vec) {
    for (int i = 0; i < vec.size(); i++) {
        if (vec[i] == vec[i + 1]) { 
            vec.remove(i);
        }
    }
}

The intent behind this function is that we could do something like this:

Vector<string> hiddenFigures = {
    "Katherine Johnson",
    "Katherine Johnson",
    "Katherine Johnson",
    "Mary Jackson",
    "Dorothy Vaughan",
    "Dorothy Vaughan"
};

deduplicate(hiddenFigures);
// hiddenFigures variable now contains ["Katherine Johnson", "Mary Jackson", "Dorothy Vaughan”]

The problem is that the above implementation of deduplicate does not work correctly. In particular, it contains three bugs. First, find these bugs by writing test cases that pinpoint potentially erroneous situations in which the provided code might fail, then explain what the problems are, and finally fix those errors in code.

Solution

There are three errors here:

  1. Calling .remove() on the Vector while iterating over it doesn’t work particularly nicely. Specifically, if you remove the element at index i and then increment i in the for loop, you’ll skip over the element that shifted into the position you were previously in.
  2. There’s an off-by-one error here: when i = vec.size() - 1, the indexing vec[i + 1] reads off the end of the Vector.
  3. The Vector is passed in by value, not by reference, so none of the changes made to it will persist to the caller.

Here are two alternate corrected versions of the code:

// solution 1
void deduplicate(Vector<string>& vec) {
    for (int i = 0; i < vec.size() - 1; ) {
        if (vec[i] == vec[i + 1]) {
            vec.remove(i);
        } else {
            i++;
        }
    }
}

// solution 2
void deduplicate(Vector<string>& vec) {
    for (int i = vec.size() - 1; i > 0; i--) {
        if (vec[i] == vec[i - 1]) {
            vec.remove(i);
        }
    }
}

3) Mirror (mirror.cpp)

Topic: Grids

Write a function ​mirror​ that accepts a reference to a ​Grid​ of integers as a parameter and flips the grid along its diagonal. You may assume the grid is square; in other words, that it has the same number of rows as columns. For example, the original Grid below would be altered to give it the new grid state shown afterwards:

Original state: 
{ { 6, 1, 9, 4},                
  {-2, 5, 8, 12},                  
  {14, 39, -6, 18},             
  {21, 55, 73, -3} }               

Mirrored state: 
 { {6, -2, 14, 21},
   {1, 5, 39, 55},
   {9, 8, -6, 73},
   {4, 12, 18, -3} }

Bonus: How would you solve this problem if the grid were not square?

Solution 
// solution
void mirror(Grid<int>& grid) {
    for (int r = 0;r < grid.numRows(); r++) {
        // start at r+1 rather than 0 to avoid double-swapping 
        for (int c = r + 1; c < grid.numCols(); c++) { 
            int temp = grid[r][c]; 
            grid[r][c] = grid[c][r]; 
            grid[c][r] = temp;
        } 
    }
}
// bonus 
void mirror(Grid<int>& grid) {
    Grid<int> result(grid.numCols(), grid.numRows());
    for (int r = 0; r < grid.numRows(); r++) {
        for (int c = 0; c < grid.numCols(); c++) {
            result[c][r] = grid[r][c];
        }
    }
    grid = result;
}

4) Check Balance (balance.cpp)

Topics: Stacks

Write a function named checkBalance that accepts a string of source code and uses a Stack to check whether the braces/parentheses are balanced. Every ( or { must be closed by a } or ) in the opposite order. Return the index at which an imbalance occurs, or -1 if the string is balanced. If any ( or { aren't closed, return the string's length.

Here are some example calls:

Code Snippet A

//   index    0123456789012345678901234567
checkBalance("if (a(4) > 9) { foo(a(2)); }") 
// returns -1 (balanced)

//   index    01234567890123456789012345678901
checkBalance("for (i=0;i<a;(3};i++) { foo{); )")
// returns 15 because } is out of order

//   index    0123456789012345678901234
checkBalance("while (true) foo(); }{ ()")
// returns 20 because } doesn't match any {

//   index    01234567
checkBalance("if (x) {")
// returns 8 because { is never closed
Solution 
int checkBalance(string code) {
    Stack<char> parens;
    for (int i = 0; i < (int) code.length(); i++) {
        char c = code[i];
        if (c == '(' || c == '{') {
        parens.push(c);
        } else if (c == ')' || c == '}') {
            if (parens.isEmpty()) {
                return i;
            }
            char top = parens.pop();
            if ((top == '(' && c != ')') || (top == '{' && c != '}')) {
                return i;
            }
        }
    }

    if (parens.isEmpty()) {
        return -1; // balanced
    } else {
        return code.length();
    }
}

5) Collection Mystery

Topics: Stacks and Queues

We're going to trace through the following code.

void collectionMystery(Stack<int>& s) { 
    Queue<int> q;
    Stack<int> s2;

    while (!s.isEmpty()) {
       if (s.peek() % 2 == 0) {
            q.enqueue(s.pop()); 
        } else {
            s2.push(s.pop());
        }
    }
    while (!q.isEmpty()) {
        s.push(q.dequeue()); 
    }
    while(!s2.isEmpty()) { 
        s.push(s2.pop());
    }
    cout<< s << endl;
}

Write the output produced by the above function when passed each of the following Stacks. Note that Stacks and Queues are written in ​front to back order, with the oldest element on the left side of the Queue/Stack.

Input Stacks:

{1, 2, 3, 4, 5, 6}                ________________________________________
{42, 3, 12, 15, 9, 71, 88}        ________________________________________
{65, 30, 10, 20, 45, 55, 6, 1}    ________________________________________
Solution 
{6, 4, 2, 1, 3, 5}
{88, 12, 42, 3, 15, 9, 71}
{6, 20, 10, 30, 65, 45, 55, 1}

6) Friend Map (friendmap.cpp)

Topics: Maps, Grids

Write a function named friendMap that takes in a Grid of strings storing friendship data and writes this data to a Map. friendMap should create and return the populated Map. The input Grid has two columns and 0 or more rows, where each location in the Grid contains a single when. If two names appear together in the same row, these two people are friends! Friendships are bi-directional, so if Sidibou is friends with Eloise, Eloise is friends with Sidibou. The Grid contains one friend relationship per row. You do not have to worry about malformed entries (Grids with more/less than two columns, empty strings, duplicates etc.)

If an input Grid looked like this:

{ {"Eloise", "Sidibou"},
  {"Sidibou", "Cheeto"} }

Then friendMap should return a Map that looks like this:

{"Sidibou":{"Eloise", "Cheeto"}, "Eloise":{"Sidibou"}, "Cheeto":{"Sidibou"}}

Here is the function definition you should implement:
Map<string, Vector<string>> friendMap(Grid<string> grid)

Solution 
Map<string, Vector<string>> friendMap(Grid<string> grid) {
    Map<string, Vector<string>> result;
    for (int row = 0; row < grid.numRows(); row++) {
        string friend1 = grid[row][0];
        string friend2 = grid[row][1];
        result[friend1].add(friend2);
        result[friend2].add(friend1);
    }
    return result;
}

7) Twice (twice.cpp)

Topics: Sets

Write a function named twice that takes a Vector of integers and returns a set containing all the numbers in the Vector that appear exactly twice.

twice({1, 3, 1, 4, 3, 7, -2, 0, 7, -2, -2, 1})
// returns {3, 7}

twice({0, 0})
// returns {0}

twice({1, 2, 3})
// returns {}

Bonus: do the same thing, but you are not allowed to declare any kind of data structure other than sets.

Solution 
// solution
Set<int> twice(Vector<int>& v) {
    Map<int, int> counts;
    for (int i : v) {
        counts[i]++;
    }
    Set<int> twice;
    for (int i : counts) {
        if (counts[i] == 2) {
            twice += i;
        }
    }
    return twice;
}

// bonus
Set<int> twice(Vector<int>& v) {
    Set<int> once;
    Set<int> twice;
    Set<int> more;
    for (int i : v) {
        if (once.contains(i)) {
            once.remove(i);
            twice.add(i);
        } else if (twice.contains(i)) {
            twice.remove(i);
            more.add(i);
        } else if (!more.contains(i)) {
            once.add(i);
        }
    }
    return twice;
}

8) Reversing a Map (reverse.cpp)

Topics: Nested data structures

Write a function

Map<int, Set<string>> reverseMap(Map<string, int>& map)

that, given a Map<string, int> that associates string values with integers, produces a Map<int, Set<string>> that’s essentially the reverse mapping, associating each integer value with the set of strings that map to it. (This is an old job interview question from 2010.)

Solution 
// One solution
Map<int, Set<string>> reverseMap(Map<string, int>& map) {
    Map<int, Set<string>> result;
    for (string oldKey : map) {
        if (!result.containsKey(map[oldKey])) {
            result[map[oldKey]] = {};
        }
        result[map[oldKey]].add(oldKey);
    }
    return result;
}

// An alternate solution, leveraging the fact that C++ auto-initializes keys
Map<int, Set<string>> result;
    for (string oldKey : map) {
        result[map[oldKey]] += oldKey;
    }
    return result;
}

9) Oh No, Big-O, Too Slow

Topics: Big-O

Give a tight bound of the nearest runtime complexity class for each of the following code snippets in Big-Oh notation, in terms of the variable N.

Code Snippet A

int sum = 0;
for (int i = 1; i <= N + 2; i++) {
    sum++;
}
for (int j = 1; j <= N * 2; j++) {
    sum++;
}
cout << sum << endl;

Code Snippet B

int sum = 0;
for (int i = 1; i <= N - 5; i++) {
    for (int j = 1; j <= N - 5; j += 2) {
        sum++;
    }
}
cout << sum << endl;

Code Snippet C

int sum = 0;
for (int i = 0; i < 1000000; i++) {
    for (int j = 1; j <= i; j++) {
        sum += N;
    }
    for (int j = 1; j <= i; j++) {
        sum += N;
    }
    for (int j = 1; j <= i; j++) {
        sum += N;
    }
}
cout << sum << endl;
Solution 
Code Snippet A has a runtime complexity of O(N).
Code Snippet B has a runtime complexity of O(N^2).
Code Snippet C has a runtime complexity of O(1).

10) More Big-0

Topics: Big-O

Below are five functions. Determine the big-O runtime of each of those pieces of code, in terms of the variable n.

void function1(int n) {
    for (int i = 0; i < n; i++) {
        cout << '*' << endl;
    }
}

void function2(int n) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << '*' << endl;
        }
    }
}

void function3(int n) {
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            cout << '*' << endl;
        }
    }
}

void function4(int n) {
    for (int i = 1; i <= n; i *= 2) {
        cout << '*' << endl;
    }
}

Bonus question, what is the big-O runtime of this function in terms of n, the number of elements in v?

int squigglebah(Vector<int>& v) {
    int result = 0;
    for (int i = 0; i < v.size(); i++) {
        Vector<int> values = v.subList(0, i);
        for (int j = 0; j < values.size(); j++) {
            result += values[j];
        }
    }
    return result;
}
Solution
  1. The runtime of this code is O(n): We print out a single star, which takes time O(1), a total of n times.
  2. The runtime of this code is O(n^2). The inner loop does O(n) work, and it runs O(n) times for a net total of O(n^2) work.
  3. This one also does O(n^2) work. To see this, note that the first iteration of the inner loop runs for n – 1 iterations, the next for n – 2 iterations, then n – 3 iterations, etc. Adding all this work up across all iterations gives (n – 1) + (n – 2) + … + 3 + 2 + 1 + 0 = O(n^2).
  4. This one runs in time O(log n). To see why this is, note that after k iterations of the inner loop, the value of i is equal to 2^k. The loop stops running when 2^k exceeds n. If we set 2^k = n, we see that the loop must stop running after k = logâ‚‚ n steps.
    Another intuition for this one: the value of i doubles on each iteration, and you can only double O(log n) times before you overtake the value n.

For the final function, let’s follow the useful maxim of "when in doubt, work inside out!"" The innermost for loop (the one counting with j) does work proportional to the size of the values list, and the values list has size equal to i on each iteration. Therefore, we can simplify this code down to something that looks like this:

int squigglebah(Vector<int>& v) {
    int result = 0;
    for (int i = 0; i < v.size(); i++) {
        Vector<int> values = v.subList(0, i);
        do O(i) work;
    }
    return result;
}

Now, how much work does it take to create the values vector? We’re copying a total of i elements from v, and so the work done will be proportional to i. That gives us this:

int squigglebah(Vector<int>& v) {
    int result = 0;
    for (int i = 0; i < v.size(); i++) {
        do O(i) work;
        do O(i) work;
    }
    return result;
}

Remember that doing O(i) work twice takes time O(i), since big-O ignores constant factors. We’re now left with this:

int squigglebah(Vector<int>& v) {
    int result = 0;
    for (int i = 0; i < v.size(); i++) {
        do O(i) work;
    }
    return result;
}

This is the same pattern as function2, and it works out to O(n^2) total time.