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question:Implement a C++ class to parse and execute mathematical expressions. The expressions should be passed to the class as strings. The expressions can contain the four basic arithmetic operations: addition, subtraction, multiplication, and division. The class should be able to handle expressions with parentheses and decimal numbers. Here's a basic class structure to get you started: ```cpp class MathExpressionParser { public: double evaluate(const std::string& expression) { // implement me } private: // add any necessary private members }; ``` Example usage of the `MathExpressionParser` class: ```cpp int main() { MathExpressionParser parser; std::cout << parser.evaluate(1 + 2 * 3) << std::endl; // outputs 7 std::cout << parser.evaluate((4 - 2) * 3) << std::endl; // outputs 6 std::cout << parser.evaluate(10 / 2) << std::endl; // outputs 5 std::cout << parser.evaluate(1.5 + 2.8) << std::endl; // outputs 4.3 return 0; } ``` Please complete the implementation of the `MathExpressionParser` class to support the described functionality.
answer:To implement the `MathExpressionParser` class, we'll use a stack-based approach to parse and evaluate the mathematical expressions. We'll also utilize the `std::istringstream` class to tokenize the input string. Here's the completed implementation: ```cpp #include <iostream> #include <string> #include <sstream> #include <stack> #include <cctype> class MathExpressionParser { public: double evaluate(const std::string& expression) { std::istringstream iss(expression); std::stack<double> values; std::stack<char> operators; while (iss.peek() != EOF) { if (std::isdigit(iss.peek()) || iss.peek() == '.') { double value; iss >> value; values.push(value); } else if (iss.peek() == '(') { iss.get(); operators.push('('); } else if (iss.peek() == ')') { iss.get(); while (operators.top() != '(') { applyOperator(values, operators); } operators.pop(); } else { char op; iss >> op; while (!operators.empty() && operators.top() != '(' && getPrecedence(op) <= getPrecedence(operators.top())) { applyOperator(values, operators); } operators.push(op); } // Consume any whitespace characters while (iss.peek() == ' ') { iss.get(); } } while (!operators.empty()) { applyOperator(values, operators); } return values.top(); } private: int getPrecedence(char op) { switch (op) { case '+': case '-': return 1; case '*': case '/': return 2; default: return 0; } } void applyOperator(std::stack<double>& values, std::stack<char>& operators) { double right = values.top(); values.pop(); double left = values.top(); values.pop(); char op = operators.top(); operators.pop(); switch (op) { case '+': values.push(left + right); break; case '-': values.push(left - right); break; case '*': values.push(left * right); break; case '/': if (right != 0) { values.push(left / right); } else { throw std::runtime_error(Division by zero); } break; } } }; ``` This implementation should support the described functionality, including parsing and executing mathematical expressions with parentheses and decimal numbers. The `applyOperator` function applies the operators in the correct order, taking into account the precedence of the operators. The `getPrecedence` function returns the precedence of a given operator. You can use the `MathExpressionParser` class as shown in your example usage. If the input expression contains any syntax errors or division by zero, the program will throw an exception.
question:Implement a Python function to find the longest common subsequence (LCS) between two strings using dynamic programming. The function should return the LCS as a string. Example usage: ```python print(find_lcs(banana, anana)) # Output: anana print(find_lcs(abcdef, zbcdfg)) # Output: bcd ``` Here's a basic function structure to get you started: ```python def find_lcs(str1, str2): # implement me pass ``` Please complete the implementation of the `find_lcs` function to find the longest common subsequence between two input strings. Note that the LCS problem is different from the longest common substring problem. The LCS can be non-contiguous, whereas the longest common substring must be contiguous. For example, the LCS of banana and anana is anana, but the longest common substring is anana as well. However, the LCS of abcdef and zbcdfg is bcd, whereas the longest common substring is only b, c, or d individually. In the case of multiple possible LCS, return any one of them.
answer:Here's the implementation of the `find_lcs` function using dynamic programming: ```python def find_lcs(str1, str2): m = len(str1) n = len(str2) dp = [[0] * (n + 1) for _ in range(m + 1)] for i in range(1, m + 1): for j in range(1, n + 1): if str1[i - 1] == str2[j - 1]: dp[i][j] = dp[i - 1][j - 1] + 1 else: dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) lcs = [] i, j = m, n while i > 0 and j > 0: if str1[i - 1] == str2[j - 1]: lcs.append(str1[i - 1]) i -= 1 j -= 1 elif dp[i - 1][j] > dp[i][j - 1]: i -= 1 else: j -= 1 return .join(reversed(lcs)) ``` This implementation first builds a dynamic programming (DP) table `dp` to store the lengths of the LCS for subproblems. The DP table is initialized with zeros, and then filled in row by row using the following recurrence relation: * If the current characters in `str1` and `str2` are equal, the length of the LCS is one more than the length of the LCS for the substrings without the current characters. * If the current characters are not equal, the length of the LCS is the maximum of the lengths of the LCS for the substrings without the current character in `str1` and without the current character in `str2`. After building the DP table, the function constructs the LCS by tracing back the path from the bottom-right corner to the top-left corner of the table. If the current characters in `str1` and `str2` are equal, the character is added to the LCS. Otherwise, the function moves in the direction of the larger value in the DP table. Finally, the function returns the LCS as a string by reversing the characters in the `lcs` list and joining them together. You can use the `find_lcs` function as shown in your example usage. Note that the function returns any one of the possible LCS in case of multiple solutions.
question:Implement a Java function to find the first duplicate in an array of integers. The function should return the first duplicate if found, or -1 if no duplicates are found. Example usage: ```java System.out.println(findFirstDuplicate(new int[] {2, 1, 3, 5, 3, 2})); // Output: 3 System.out.println(findFirstDuplicate(new int[] {2, 4, 3, 5, 1})); // Output: -1 ``` Here's a basic function structure to get you started: ```java public class Main { public static int findFirstDuplicate(int[] arr) { // implement me return -1; } public static void main(String[] args) { System.out.println(findFirstDuplicate(new int[] {2, 1, 3, 5, 3, 2})); // Output: 3 System.out.println(findFirstDuplicate(new int[] {2, 4, 3, 5, 1})); // Output: -1 } } ``` Please complete the implementation of the `findFirstDuplicate` function to find the first duplicate in the input array. Note that the function should return the first duplicate value, not its index. If there are multiple duplicates, return the first one that appears in the array. If no duplicates are found, return -1. You can assume that the input array is not null and contains at least one element. You are allowed to use additional data structures, such as sets or maps, to help with the implementation.
answer:Here's the implementation of the `findFirstDuplicate` function using a `HashSet` to keep track of the elements we've seen so far: ```java import java.util.HashSet; public class Main { public static int findFirstDuplicate(int[] arr) { HashSet<Integer> seen = new HashSet<>(); for (int num : arr) { if (!seen.add(num)) { return num; } } return -1; } public static void main(String[] args) { System.out.println(findFirstDuplicate(new int[] {2, 1, 3, 5, 3, 2})); // Output: 3 System.out.println(findFirstDuplicate(new int[] {2, 4, 3, 5, 1})); // Output: -1 } } ``` This implementation works by iterating over the input array and adding each element to the `seen` set. If an element is already present in the set (i.e., `add` returns `false`), it means we've found a duplicate, and we return that element immediately. If we reach the end of the array without finding any duplicates, we return -1. Using a `HashSet` allows us to keep track of the elements we've seen so far in O(1) time, making the overall time complexity of the function O(n), where n is the length of the input array. The space complexity is also O(n) because in the worst case, we might need to store all elements in the set.
question:Rewrite the rules of a game in ten words or less. Monopoly
answer:: Buy property, collect rent, bankrupt opponents. Chess: Capture the king, protect your own. UNO: Match colors, numbers, and special cards to win. Pictionary: Draw and guess words to advance on the board. Catan: Collect resources, build settlements, gain victory points. Clue: Solve the murder mystery by eliminating suspects, weapons, and rooms. Risk: Conquer the world, defeat enemies, and defend territories. Twister: Spin and place body parts on colored dots. Scrabble: Make words with letter tiles to score points. Jenga: Remove blocks, don't topple the tower.