More Practice Problems

Written by Clinton Kwarteng, drawing upon materials from previous quarters.

These problems will give you more practice with concepts covered on the mid-quarter exam. Note that some of these questions might involve topics we have not yet covered in class.

Programming in C++

1. References Available Upon Request

Topic: Reference parameters, range-based for loops

Reference parameters are an important part of C++ programming, but can take some getting used to if you’re not familiar with them. Trace through the following code. What does it print?

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. Pig-Latin

Topics: Strings, reference parameters, return types

Write two functions, pigLatinReturn and pigLatinReference, that accept a string and convert said string into its pig-Latin form. To convert a string into pig-Latin, you must follow these steps:

  • Split the input string into 2 strings: a string of characters BEFORE the first vowel, and a string of characters AFTER (and including) the first vowel.
  • Append the first string (letters before the first vowel) to the second string.
  • Append the string "ay" to the resulting string.

Here are a few examples…

nick -> icknay

chase -> asechay

chris -> ischray

You will need to write this routine in two ways: once as a function that returns the pig-Latin string to the caller, and once as a function that modifies the supplied parameter string and uses it to store the resulting pig-Latin string. These will be done in pigLatinReturn and pigLatinReference, respectively. You may assume that your input is always a one-word, all lowercase string with at least one vowel.

Here's a code example of how these functions differ…

string name = "julie";
string str1 = pigLatinReturn(name);
cout << str1 << endl; // prints "uliejay"

pigLatinReference(name);
cout << name << endl; // prints "uliejay"

Once you've written these functions, discuss with your section the benefits and drawbacks of these two approaches. Which do you feel is easier to write? Which do you think is more convenient for the caller? Do you think one is better style than the other?

Solution 
// Use const because VOWELS won't change -- no need to declare repeatedly
// in isVowel.
const string VOWELS = "aeiouy";

// Helper function, which I'd highly recommend writing!
bool isVowel(char ch) {
    // A little kludgy, but the handout guarantees that
    // ch will ALWAYS be lower case :)
    // NOTE: For an assignment, you probably want a more robust isVowel.
    return VOWELS.find(ch) != string::npos;
}

string pigLatinReturn(string input) {
    int strOneIndex = 0;
    for (int i = 0; i < input.length(); i++) {
        if (isVowel(input[i])) {
            strOneIndex = i;
            break;
        }
    }
    string strOne = input.substr(0, strOneIndex);
    string strTwo = input.substr(strOneIndex);
    return strTwo + strOne + "ay";
}

void pigLatinReference(string &input) {
    int strOneIndex = 0;
    for (int i = 0; i < input.length(); i++) {
        if (isVowel(input[i])) {
            strOneIndex = i;
            break;
        }
    }
    string strOne = input.substr(0, strOneIndex);
    string strTwo = input.substr(strOneIndex);
    input = strTwo + strOne + "ay";
}

Notice how similar these two approaches are – the only difference is how the result is handled at the very end. To address the discussion questions, although the pigLatinReference function is marginally more efficient because it doesn't need to make a copy of the input string, pigLatinReturn is probably more intuitive for both the caller and the writer: if the function's job is to somehow output some product, returning is the most explicit way to do so. In that way, a function that returns is also better style – it's makes the purpose of the function clearer to the reader.

If you wanted to combine the efficiency of pigLatinReference with the clarity of pigLatinReturn, I would recommend writing a function that takes in the input string by const reference, basically

string pigLatin(const string &input);

Although the const isn't explicitly necessary, it's nice to have because you never need to modify input. Moreover, you still get the efficiency gains from pass-by-reference while also writing very-understandable code.

ADTs

1. Mirror

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 grid below that comes first 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[r][c] = grid[c][r];
        }
    }
    grid = result;
}

2. Check Balance

Topic: 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 { are never closed, return the string's length.

Here are some example calls:

//   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 < 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
    } 
    return code.length();
}

3. Collection Mystery

Topics: Stacks, queues

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.

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}

4. Twice

Topic: 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.

Example: passing {1, 3, 1, 4, 3, 7, -2, 0, 7, -2, -2, 1} returns {3, 7}.

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;
}

5. Reversing a Map

Topic: 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

Here’s one possible implementation.

Map<int, Set<string>> reverseMap(Map<string, int>& map) {
    Map<int, Set<string>> result;
    for (string oldKey : map) {
        // Note: we check containsKey here but this isn't 
        // necessary since Maps auto-initialize values 
        // on square bracket access if the key is not 
        // present
        if (!result.containsKey(map[oldKey])) {
            result[map[oldKey]] = {};
        }
        result[map[oldKey]].add(oldKey);
    }
    return result;
}

6. Stack vs Vector showdown

Topics: Stacks, vectors

We will analyze a simple program that emulates a text editor. The program accepts input from a user which can be of two forms:

  • ADD string: This adds a one word string into the text editor

  • UNDO: This removes the most recently added string from the editor.

When the user finishes providing input (by hitting ENTER), we print a string which represents the words in the text editor, in the order they were added.

Example: If we receive user input like "ADD I", "ADD Love", "ADD 106A", "UNDO", "ADD 106B", the program prints out "I love 106B".

int main() {
    Stack<string> words;

    while (true) {
        string command = getLine("Enter a command (press enter to quit): ");
        if (command == "") {
            break;
        }

        Vector<string> splitCommands = stringSplit(command, " ");
        if (splitCommands[0] == "ADD") {
            words.push(splitCommands[1]);
        } else if (splitCommands[0] == "UNDO") {
            words.pop();
        }
    }

    string result;
    while (!words.isEmpty()) {
        result = words.pop() + result;
    }
    cout << result << endl;
    return 0;
}

7. Maps

Topic: Maps

Here, we analyze a simple program that uses maps to count elements in a vector

Map<string, int> countElements(Vector<string>& names) {
    Map<string, int> frequency;
    for (string name: names) {
        // When using the square bracket syntax([]), we don't need to initialize
        // names in the map before adding unto it. If name doesn't already exist
        // in the map, [] in the Stanford Map will initialize it for us.
        frequency[name] += 1;
    }
    return frequency;
}

Big-O

1. Oh No, Big-O, Too Slow

Topics: Big-O, code analysis, ADTs

What is the Big O runtime of the following functions, in terms of the variable N.

Code Snippet A

Vector<int> v;
for (int i = 1; i <= N + 2; i++) {
    v.add(i);
}

for (int i = 1; i < N; I ++) {
    v.insert(0, i);
}

Code Snippet B

Set<string> s;
for (int i = 1; i <= N - 5; i++) {
    s.add(i);
}

Code Snippet C

Map<int, int> m;
for (int i = 1; i <= 3*N; i++) {
    m[i] = i * 4;
}

Code Snippet D

Stack<int> s;
for (int i = 1; i <= N; i++) {
    s.push(i)
}

for (int i = 1; i <= N/2; i++) {
    s.pop(i)
}
Solution 
Code Snippet A has a runtime complexity of O(N^2): The first for loop runs in 
O(N), because adding to a vector is constant time O(1), and we add N time. The
second for loop runs N times, but inside this loop we perform an O(N) 
operation, which is inserting at the front of a vector. Therefore Big O of 
second loop is O(N^2). This makes O(N) + O(N^2) = O(N^2), because we only pick 
the highest order Big O.

Code Snippet B has a runtime complexity of O(NlogN): The first for loop runs in 
N. Inside of the loop, we perform an O(logN) operation, which is adding to a set. 
This makes O(N * logN) = O(NlogN)

Code Snippet C has a runtime complexity of O(NlogN): This is very similar to B 
above. Inserting in a map also runs in O(logN). If we do that N times, we get 
O(NlogN).

Code Snippet D has a runtime complexity of O(N):  Pushing unto a stack runs in 
O(1) time, so the first for loop runs in O(N). Popping off a stack runs in O(1)
time so the second for loop runs in O(N/2) = O(N). In total, O(N) + O(N) = O(N), 
since we throw away multipliers.

2. More Big-O

Topics: Big-O, code analysis, ADTs

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;
    }
}

Finally, 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 = log₂ 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 in the previous problem, and it works out to O(n^2) total time.

Recursion

1. Recursion Mystery Part 1

Topics: Recursive function calls, return value tracing

Code Snippet A

int recursionMystery(int x, int y) {
    if (x < y) {
        return x;
    } else {
        return recursionMystery(x - y, y);
    }
}

For each call to the above recursive function, indicate what value is returned by the function call.

Call                                   Return value  

recursionMystery(6, 13);               ______________
recursionMystery(14, 10);              ______________
recursionMystery(37, 12);              ______________
Solution

6

4

1

2. Recursion Mystery Part 2

Topics: Recursive function calls, output tracing

void recursionMystery2(int x, int y) {
    if (y == 1) { 
        cout << x;
    } else {
        cout << (x * y) << ", ";
        recursionMystery2(x, y - 1);
        cout << ", " << (x * y);
    } 
}

For each call to the above recursive function, write the output that would be produced, as it would appear on the console.

Call                                   Output
        
recursionMystery2(4, 1);        ___________________________________
recursionMystery2(4, 2);        ___________________________________
recursionMystery2(8, 2);        ___________________________________
recursionMystery2(4, 3);        ___________________________________
recursionMystery2(3, 4);        ___________________________________
Solution

4

8, 4, 8

16, 8, 16

12, 8, 4, 8, 12

12, 9, 6, 3, 6, 9, 12

3. Recursion Tracing

Topics: Recursion, strings, recursion tracing

Below is a recursive function to reverse a string.

string reverseOf(string s) {
    if (s.empty()) {
        return "";
    } else {
        return reverseOf(s.substr(1)) + s[0];
    }
}

Trace through the execution of reverseOf("stop") along the lines of what we did in lecture, showing recursive call information for each call that’s made and how the final value gets computed.

Solution

Our initial call to reverseOf("stop") fires off a call to reverseOf("top"). This call fires off a call to reverseOf("op"). This in turn calls reverseOf("p"). This in turn calls reverseOf(""). This triggers the base case and returns the empty string. (Notice that the reverse of the empty string "" is indeed the empty string ""). We now append p to return "p". We now append o to return "po". We append t to return "pot". And finally we append s to return "pots" back to whoever called us. Yay!

4. Sum of Squares

Topics: Recursion

Write a recursive function named sumOfSquares that takes an int n and returns the sum of squares from 1 to n. For example, sumOfSquares(3)should return 1^2 + 2^2 + 3^2 = 14. If n is negative, you should report an error to that effect.

Solution 
int sumOfSquares(int n) {
    if (n < 0) {
        error("Value of provided n was negative");
    } else if (n == 0) {
        return 0;
    } else {
        return n * n + sumOfSquares(n-1);
    }
}

5. Zig Zag

Topics: Recursion, printing output to console

Write a recursive function named zigzag that returns a string of n characters as follows. The middle character (or middle two characters if n is even) is an asterisk (*). All characters before the asterisks are '<'. All characters after are '>'. Report an error if n is not positive.

Call            Output
zigzag(1)       * 
zigzag(4)       <**> 
zigzag(9)       <<<<*>>>>
Solution 
string zigzag(int n) {
    if (n < 1) {
        error("The value of n was negative");
    } else if (n == 1) {
        return "*";
    } else if (n == 2) {
        return "**";
    } else {
        return "<" + zigzag(n-2) + ">";
    }
}

6. Double Stack

Topics: Recursion, stacks

Write a recursive function named doubleStack that takes a reference to a stack of ints and replaces each integer with two copies of that integer. For example, if s stores {1, 2, 3}, then doubleStack(s) changes it to {1, 1, 2, 2, 3, 3}.

Solution 
void doubleStack(Stack<int>& s) 
{ 
    if (!s.isEmpty()) {
        int n = s.pop();
        doubleStack(s); 
        s.push(n); 
        s.push(n);
    }
}

7. String Subsequences

Topics: Recursion, verifying properties

Write a recursive function named isSubsequence that takes two strings and returns true if the second string is a subsequence of the first string. A string is a subsequence of another if it contains the same letters in the same order, but not necessarily consecutively. You can assume both strings are already lower-cased.

Call                                    Output
isSubsequence("computer", "core")       false 
isSubsequence("computer", "cope")       true 
isSubsequence("computer", "computer")   true
Solution 
bool isSubsequence(string big, string small) 
{ 
    if (small.empty()) {
        return true;
    } else if (big.empty()) {
        return false;
    } else {
        if (big[0] == small[0]) {
            return isSubsequence(big.substr(1), small.substr(1));
        } else {
            return isSubsequence(big.substr(1), small);
        } 
    }
}

8. Recursion and Time Complexity and Exponents, (Big-)O My

Topics: Recursion, time complexity, Big-O, algorithm comparison

Below is a simple function that computes the value of m^n when n is a nonnegative integer:

int raiseToPower(int m, int n) {
    int result = 1;
    for (int i = 0; i < n; i++) {
        result *= m;
    }
    return result;
}

1) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

2) If it takes 1 microsecond (μs) to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

Below is a recursive function that computes the value of m^n when n is a nonnegative integer:

int raiseToPower(int m, int n) {
    if (n == 0) return 1;
    return m * raiseToPower(m, n - 1);
}

3) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

4) If it takes 1μs to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

Here’s an alternative recursive function for computing m^n that uses a technique called exponentiation by squaring. The idea is to modify the recursive step as follows:

  • If n is an even number, then we can write n as n = 2k. Then m^n = m^(2k) = (m^k)^2.
  • If n is an odd number, then we can write n as n = 2k + 1. Then m^n = m^(2k+1) = m * m^(2k) = m * (m^k)^2.

Based on this observation, we can write this recursive function:

int raiseToPower(int m, int n) {
    if (n == 0) {
        return 1;
    } else if (n % 2 == 0) {
        int halfPower = raiseToPower(m, n / 2);
        return halfPower * halfPower;
    } else {
        int halfPower = raiseToPower(m, n / 2);
        return m * halfPower * halfPower;
    }
}

5) What is the big-O complexity of the above function, written in terms of m and n? You can assume that it takes time O(1) to multiply two numbers.

6) If it takes 1μs to compute raiseToPower(100, 100), approximately how long will it take to compute raiseToPower(200, 10000)?

Solution

  1. This function runs in time O(n). It runs the loop n times, at each step doing O(1) work. There is no dependence on m in the runtime.

  2. We know that this code runs in time O(n), so it scales roughly linearly with the size of n. Therefore, if it took 1μs to compute a value when n = 100, it will take roughly 100 times longer when we plug in n = 10000. As a result, we’d expect this code would take about 100μs to complete.

  3. If we trace through the recursion, we’ll see that we make a total of n recursive calls, each of which is only doing O(1) work. Adding up all the work done by these recursive calls gives us a total of O(n) work, as before.

  4. As before, this should take about 100μs.

  5. Notice that each recursive call does O(1) work (there are no loops anywhere here), then calls itself on a problem that’s half as big as the original one. This means that only O(log n) recursive calls will happen (remember that repeatedly dividing by two is the hallmark of a logarithm), so the total work done here is O(log n).

  6. We know that the runtime when n = 100 is roughly 1μs. Notice that 100^2 = 10,000, so we’re essentially asking for the runtime of this function when we square the size of the input. Also notice that via properties of logarithms that log n^2 = 2 log n. Therefore, since we know the runtime grows roughly logarithmically and we’ve squared the value of n, this should take about twice as long as before, roughly 2μs.

Backtracking

1. Longest Common Subsequence

Topic: Recursive Backtracking

Write a recursive function named longestCommonSubsequence that returns the longest common subsequence of two strings passed as arguments. Some example function calls and return values are shown below.

Recall that if a string is a subsequence of another, each of its letters occurs in the longer string in the same order, but not necessarily consecutively.

Hint: In the recursive case, compare the first character of each string. What one recursive call can you make if they are the same? What two recursive calls do you try if they are different?

longestCommonSubsequence("cs106a", "cs106b") --> "cs106" 
longestCommonSubsequence("nick", "julie") --> "i" 
longestCommonSubsequence("karel", "c++") --> "" 
longestCommonSubsequence("she sells", "seashells") --> "sesells"
Solution 
string longestCommonSubsequence(string s1, string s2) {
	if (s1.length() == 0 || s2.length() == 0) {
		return "";
 	} else if (s1[0] == s2[0]) {
 		return s1[0] + longestCommonSubsequence(s1.substr(1), 
 							s2.substr(1));
 	} else {
 		string choice1 = longestCommonSubsequence(s1, s2.substr(1));
		string choice2 = longestCommonSubsequence(s1.substr(1), s2);
		if (choice1.length() >= choice2.length()) {
			return choice1;
		} else {
			return choice2;
		}
	}
}

2. Domino Tiling

Topic: Recursive backtracking

Imagine you have a 2 × n grid that you’d like to cover using 2 × 1 dominoes. The dominoes need to be completely contained within the grid (so they can’t hang over the sides), can’t overlap, and have to be at 90° angles (so you can’t have diagonal or tilted tiles). There’s exactly one way to tile a 2 × 1 grid this way, exactly two ways to tile a 2 × 2 grid this way, and exactly three ways to tile a 2 × 3 grid this way (can you see what they are?) Write a recursive function

int numWaysToTile(int n)

that, given a number n, returns the number of ways you can tile a 2 × n grid with 2 × 1 dominoes.

Solution

If you draw out a couple of sample tilings, you might notice that every tiling either starts with a single vertical domino or with two horizontal dominoes. That means that the number of ways to tile a 2 × n (for n ≥ 2) is given by the number of ways to tile a 2 × (n – 1) grid (because any of them can be extended into a 2 × n grid by adding a vertical domino) plus the number of ways to tile a 2 × (n – 2) grid (because any of them can be extended into a 2 × n grid by adding two horizontal dominoes). From there the question is how to compute this. You could do this with regular recursion, like this:

int numWaysToTile(int n) {
    /* There's one way to tile a 2 x 0 grid: put down no dominoes. */
    if (n == 0) return 1;

    /* There's one way to tile a 2 x 1 grid: put down a vertical domino. */
    if (n == 1) return 1;
    
    /* Recursive case: Use the above insight. */
    return numWaysToTile(n - 1) + numWaysToTile(n - 2);
}

3. Cracking Passwords

Topic: Recursive backtracking

Write a function crack that takes in the maximum length a site allows for a user's password and tries to find the password into an account by using recursive backtracking to attempt all possible passwords up to that length (inclusive). Assume you have access to the function bool login(string password) that returns true if a password is correct. You can also assume that the passwords are entirely alphabetic and case-sensitive. You should return the correct password you find, or the empty string if you cannot find the password. You should return the empty string if the maximum length passed is 0 and raise an error if the length is negative.

Security note: The ease with which computers can brute-force passwords is the reason why login systems usually permit only a certain number of login attempts at a time before timing out. It’s also why long passwords that contain a variety of different characters are better! Try experimenting with how long it takes to crack longer and more complex passwords. See the comic here for more information: https://xkcd.com/936/

Solution 
string crackHelper(string soFar, int maxLength) {
	if (login(soFar)) {
		return soFar;
	}
	if (soFar.size() == maxLength) {
		return "";
	}
	for (char c = 'a'; c <= 'z'; c++) {
		string password = crackHelper (soFar + c, maxLength);
		if (password != "") {
			return password;
		}
 		// Also check uppercase
 		char upperC = toupper(c);
 		password = crackHelper (soFar + upperC, maxLength);
		if (password != "") {
 			return password;
 		}
 	}
	return "";
}

string crack(int maxLength) {
	if (maxLength < 0) {
 		error("max length cannot be negative!);";
	}
 	return crackHelper("", maxLength);
}

4. Change We Can Believe In

Topic: Recursive backtracking

In the US, as is the case in most countries, the best way to give change for any total is to use a greedy strategy – find the highest-denomination coin that’s less than the total amount, give one of those coins, and repeat. For example, to pay someone 97¢ in the US in cash, the best strategy would be to

  • give a half dollar (50¢ given, 47¢ remain), then
  • give a quarter (75¢ given, 22¢ remain), then
  • give a dime (85¢ given, 12¢ remain), then
  • give a dime (95¢ given, 2¢ remain), then
  • give a penny (96¢ given, 1¢ remain), then
  • give another penny (97¢ given, 0¢ remain).

This uses six total coins, and there’s no way to use fewer coins to achieve the same total.

However, it’s possible to come up with coin systems where this greedy strategy doesn’t always use the fewest number of coins. For example, in the tiny country of Recursia, the residents have decided to use the denominations 1¢, 12¢, 14¢, and 63¢, for some strange reason. So suppose you need to give back 24¢ in change. The best way to do this would be to give back two 12¢ coins. However, with the greedy strategy of always picking the highest-denomination coin that’s less than the total, you’d pick a 14¢ coin and ten 1¢ coins for a total of eleven coins. That’s pretty bad!

Your task is to write a function

int fewestCoinsFor(int cents, Set<int>& coins)

that takes as input a number of cents and a Set indicating the different denominations of coins used in a country, then returns the minimum number of coins required to make change for that total. In the case of US coins, this should always return the same number as the greedy approach, but in general it might return a lot fewer!

You can assume that the set of coins always contains a 1¢ coin, so you never need to worry about the case where it’s simply not possible to make change for some total. You can also assume that there are no coins worth exactly 0¢ or a negative number of cents. Finally, you can assume that the number of cents to make change for is nonnegative.

Makes cents? I certainly hope so.

Solution

The idea behind both solutions is the following: if we need to make change for zero cents, the only (and, therefore, best!) option is to use 0 coins. Otherwise, we need to give back at least one coin. What’s the first coin we should hand back? We don’t know which one it is, but we can say that it’s got to be one of the coins from our options and that that coin can’t be worth more than the total. So we’ll try each of those options in turn, see which one ends up requiring the fewest coins for the remainder, then go with that choice. The code for this is really elegant and is shown here:

SOLUTION 1

/**
 * Given a collection of denominations and an amount to give in change, returns
 * the minimum number of coins required to make change for it.
 *
 * @param cents How many cents we need to give back.
 * @param coins The set of coins we can use.
 * @return The minimum number of coins needed to make change.
 */
int fewestCoinsFor(int cents, Set<int>& coins) {
	/* Base case: You need no coins to give change for no cents. */
	if (cents == 0) {
		return 0;
	}
	/* Recursive case: try each possible coin that doesn’t exceed the
	* total as as our first coin.
	*/
	else {
		int bestSoFar = cents + 1; // Can never need this many coins;

		/* What is this for loop doing? We use this loop to explore 
		* all the options of coins we have. To make the change, all of 
		* the coins in our set are valid options, so we make recursive
		* calls, each of which chooses a coin. It is very similar to 
		* the subset problems we've done in class, where we have two 
		* choices(either add or don't add to the set). Here, we either
		* add coin1 or coin2, ..., coinN.
		*/
		for (int coin: coins) {
			// If this coin doesn’t exceed the total, try using it.
			if (coin <= cents) {
				/* Use this coin in the change. We add 1 because 
				we've used one coin
				*/
				int numCoinsUsed = 1 + 
					fewestCoinsFor(cents - coin, coins);
				/* We want to find the fewest number of coins, 
				* so compare number of coins used when we use
				* the current coin to the minimum number of 
				* coins so far.
				*/
				bestSoFar = min(bestSoFar, numCoinsUsed);
			}
		}
		return bestSoFar;
 	}
}

5. Weights and Balances

Topics: Recursive backtracking

I am the only child of parents who weighed,
measured, and priced everything; for whom
what could not be weighed, measured, and
priced had no existence.
—Charles Dickens, Little Dorrit, 1857


In Dickens’s time, merchants measured many commodities using weights and a two-pan balance – a practice that continues in many parts of the world today. If you are using a limited set of weights, however, you can only measure certain quantities accurately.

For example, suppose that you have only two weights: a 1-ounce weight and a 3-ounce weight. With these you can easily measure out 4 ounces, as shown below:

two-pan balance with unmarked weight on the left side and two weights on the right side, marked 1 and 3. The balance is level.

It’s more interesting to discover that you can also measure out 2 ounces by shifting the 1-ounce weight to the other side, as follows below:

two-pan balance with weight marked 1 and unmarked weight on the left side and one weight on the right side, marked 3. The balance is level.

Write a recursive function

bool isMeasurable(int target, Vector<int>& weights)

that determines whether it is possible to measure out the desired target amount with a given set of weights, which is stored in the vector weights.

As an example, the function call

Vector<int> weights = {1, 3};
isMeasurable(2, weights);

should return true because it is possible to measure out two ounces using the sample weight set as illustrated in the preceding diagram. On the other hand, calling

Vector<int> weights = {1, 3};
isMeasurable(5, weights);

should return false because it is impossible to use the 1- and 3-ounce weights to add up to 5 ounces. However, the call

Vector<int> weights = {1, 3, 7};
isMeasurable(6, weights);

should return true: you can measure the six-ounce weight by placing it and the one-ounce weight on one side of the scale and by placing the seven-ounce weight on the other.

Here’s a function question to ponder: let’s say that you get to choose n weights. Which ones would you pick to give yourself the best range of weights that you’d be capable of measuring?

Solution

Imagine that we start off by putting the amount to be measured (call it n) on the left side of the balance (as shown in the picture on the handout). This makes the target of measurement equal to n. Imagine that there is some way to reach this target n. If we put the weights on the scale one at a time, we can look at where we put that first weight (let’s suppose it weighs w). It must either

  • go on the left side, making the target weight on the scale n + w,
  • go on the right side, making the target weight on the scale n - w

We've seen the first two options before, the classic with and without. In this question, there's a third option too:

  • not get used at all, leaving the target weight on the scale n. To see why this is the case, consider a case where you're measuring something weighing 0 ounces. You'd not need any weight to measure 0 ounces so you don't have to put anything on the scale.

If it is indeed truly possible to measure n, then one of these three options has to be the way to do it, even if we don’t know which one it is. The question we then have to ask is whether it’s then possible to measure the new net imbalance using the weights that remain – which we can determine recursively! On the other hand, if it’s not possible to measure n, then no matter which option we choose, we’ll find that there’s no way to use the remaining weights to make everything balanced!

If we’re proceeding recursively, which we are here, we need to think about our base case. There are many options we can choose from. One simple one is the following: imagine that we don’t have any weights at all, that we’re asked to see whether some weight is measurable using no weights. In what circumstances can we do that? Well, if what we’re weighing has a nonzero weight, we can’t possibly measure it – placing it on the scale will tip it to some side, but that doesn’t tell us how much it weighs. On the other hand, if what we’re weighing is completely weightless, then putting it on the scale won’t cause it to tip, convincing us that, indeed, it is weightless! So as our base case, we’ll say that when we’re down to no remaining weights, we can measure n precisely if n = 0. With that in mind, here’s our code:

bool isMeasurable(int target, Vector<int>& weights) {
	if (weights.isEmpty()) {
		return target == 0; // base case; no weights left to place
	} else {
		int last = weights[weights.size() - 1]; //using last index is fastest
		weights.remove(weights.size() - 1);

		// "choose and explore" all of the three possibilities
		// This exactly matches the analysis we did above.
		bool weightOnLeftSide = isMeasurable(target + last, weights);
		bool weightOnRightSide = isMeasurable(target - last, weights);
		bool ignoreWeight = isMeasurable(target, weights);
		// un-choose
		weights.add(last);
		// One of these three options will work if we can measure
		// the target weight.
		return weightOnLeftSide || weightOnRightSide || ignoreWeight;
 	}
}