C++ Online Test

A playlist is considered a repeating playlist if any of the songs contain a pointer to a previous song in the playlist. Otherwise, the playlist will end with the last song which points to NULL.

Implement a function isInRepeatingPlaylist that, efficiently with respect to time used, returns true if a playlist is repeating or false if it is not.

For example, the following code prints "true" as both songs point to each other.

Song* first = new Song("Hello");
Song* second = new Song("Eye of the tiger");
    
first->next(second);
second->next(first);

std::cout << std::boolalpha << first->isInRepeatingPlaylist();

As part of a data processing pipeline, complete the implementation of the makePipeline method:

• The makePipeline method should accept a vector of functions, and it should return a new function that accepts one parameter arg.
• The returned function should call the first function in makePipeline with the parameter arg, and call the second function with the result of the first function.
• The returned function should continue calling each function in makePipeline in order, following the same pattern, and return the value from the last function.

For example,

std::vector<std::function<int (int)>> functions;
functions.push_back([] (int x) -> int { return x * 3; });
functions.push_back([] (int x) -> int { return x + 1; });
functions.push_back([] (int x) -> int { return x / 2; });
std::function<int (int)> func = makePipeline(functions);

and then calling func(3) should return 5.

Implement function countNumbers that accepts a sorted vector of unique integers and, efficiently with respect to time used, counts the number of vector elements that are less than the parameter lessThan.

For example, for vector v containing { 1, 3, 5, 7 }, countNumbers(v, 4) should return 2 because there are two vector elements less than 4.

Implement the unique_names method. When passed two vectors of names, it will return a vector containing the names that appear in either or both vectors. The returned vector should have no duplicates.

For example, calling unique_names(std::vector<std::string>{"Ava", "Emma", "Olivia"}, std::vector<std::string>{"Olivia", "Sophia", "Emma"}) should return a vector containing Ava, Emma, Olivia, and Sophia in any order.

Multiple choice test has several multiple choice questions. Each question can have only one correct answer. Additionally, timed multiple choice test can specify the time allowed for solving each question in the test.

The code below satisfies this specification, but the customer complained that the memory usage of the program constantly increases. Fix this problem.

User interface contains two types of user input controls: TextInput, which accepts all characters and NumericInput, which accepts only digits.

Implement the following methods:

• add on class TextInput - adds the given character to the current value
• getValue on class TextInput - returns the current value
• add on class NumericInput - overrides the base class method so that each non-numeric character is ignored

For example, the following code should output "10":

TextInput* input = new NumericInput();
input->add('1');
input->add('a');
input->add('0');
std::cout << input->getValue();

As a part of the route planner, the routeExists function is used as a quick filter if the destination is reachable, before using more computationally intensive procedures for finding the optimal route.

The roads on the map are rasterized and produce a matrix of boolean values - true if the road is present or false if it is not. The roads in the matrix are connected only if the road is immediately left, right, below or above it.

Finish the routeExists function so that it returns true if the destination is reachable or false if it is not. The fromRow and fromColumn parameters are the starting row and column in the mapMatrix. The toRow and toColumn are the destination row and column in the mapMatrix. The mapMatrix parameter is the above mentioned matrix produced from the map.

For example, for the given rasterized map, the code below should return true since the destination is reachable:

std::vector<std::vector<bool>> mapMatrix {
	{true, false, false},
	{true, true, false},
	{false, true, true}
};

routeExists(0, 0, 2, 2, mapMatrix);

Implement the function findRoots to find the roots of the quadratic equation: ax² + bx + c = 0.  The function should return a pair containing both roots in any order. If the equation has only one solution, the function should return that solution as both elements of the pair. The equation will always have at least one solution.

The roots of the quadratic equation can be found with the following formula:

For example, findRoots(2, 10, 8) should return a pair containing (-1, -4) or (-4, -1) as the roots of the equation 2x² + 10x + 8 = 0 are -1 and -4.

Write a function that, when passed a list and a target sum, returns, efficiently with respect to time used, two distinct zero-based indices of any two of the numbers, whose sum is equal to the target sum. If there are no two numbers, the function should return (-1, -1).

For example, findTwoSum({ 3, 1, 5, 7, 5, 9 }, 10) should return a std::pair<int, int> containing any of the following pairs of indices:

• 0 and 3 (or 3 and 0) as 3 + 7 = 10
• 1 and 5 (or 5 and 1) as 1 + 9 = 10
• 2 and 4 (or 4 and 2) as 5 + 5 = 10

A TrainComposition is built by attaching and detaching wagons from the left and the right sides, efficiently with respect to time used.

For example, if we start by attaching wagon 7 from the left followed by attaching wagon 13, again from the left, we get a composition of two wagons (13 and 7 from left to right). Now the first wagon that can be detached from the right is 7 and the first that can be detached from the left is 13.

Implement a TrainComposition that models this problem.

Binary search tree (BST) is a binary tree where the value of each node is larger or equal to the values in all the nodes in that node's left subtree and is smaller than the values in all the nodes in that node's right subtree.

Write a function that, efficiently with respect to time used, checks if a given binary search tree contains a given value.

For example, for the following tree:

• n1 (Value: 1, Left: null, Right: null)
• n2 (Value: 2, Left: n1, Right: n3)
• n3 (Value: 3, Left: null, Right: null)

Call to contains(n2, 3) should return true since a tree with root at n2 contains number 3.