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question:Write a program in C++ that determines if a string contains any duplicate characters. The string contains only lowercase letters. Return true if string has duplicate characters and false otherwise. Optimize the algorithm as much as possible. Here is some sample data: Input: abcde Output: false Input: abcdea Output: true
answer:Here's an optimized C++ function that uses a `std::unordered_set` to keep track of unique characters in the string: ```cpp #include <unordered_set> #include <string> bool hasDuplicateCharacters(const std::string& str) { std::unordered_set<char> uniqueChars; for (char c : str) { if (!uniqueChars.insert(c).second) { return true; } } return false; } ``` This function iterates through each character in the input string and attempts to insert it into the `uniqueChars` set. If the insertion fails (i.e., the character is already in the set), the function immediately returns `true`, indicating the presence of a duplicate character. If the loop completes without finding any duplicates, the function returns `false`. This algorithm has a time complexity of O(n), where n is the length of the input string, since `std::unordered_set` operations (insertion and lookup) typically have an average time complexity of O(1). You can use this function like this: ```cpp int main() { std::string str1 = abcde; std::string str2 = abcdea; std::cout << std::boolalpha << hasDuplicateCharacters(str1) << std::endl; // Output: false std::cout << std::boolalpha << hasDuplicateCharacters(str2) << std::endl; // Output: true return 0; } ``` This function is optimized for performance, as it uses a data structure (`std::unordered_set`) that provides fast lookup and insertion operations. It also returns as soon as it finds a duplicate character, which reduces unnecessary iterations.
question:Implement the above approach using a bool array of size 26 where each index represents a lowercase English letter. Here is some sample data: Input: abcde Output: false Input: abcdea Output: true Secondly, using the above string, how would you find out if a string contains all lowercase English letters at least once or not.
answer:Here's the C++ function that uses a `bool` array of size 26 to detect duplicate characters in a string: ```cpp #include <string> bool hasDuplicateCharacters(const std::string& str) { bool seen[26] = {false}; for (char c : str) { int index = c - 'a'; if (seen[index]) { return true; } seen[index] = true; } return false; } ``` In this function, we create a `bool` array `seen` of size 26, where each index corresponds to a lowercase English letter (e.g., `seen[0]` represents 'a', `seen[1]` represents 'b', and so on). We then iterate through each character `c` in the input string, calculate its corresponding index in the `seen` array using `c - 'a'`, and check if the character has been seen before. If it has, we immediately return `true`, indicating the presence of a duplicate character. If the loop completes without finding any duplicates, we return `false`. To find out if a string contains all lowercase English letters at least once, we can modify the above approach as follows: ```cpp #include <string> bool containsAllLowercaseLetters(const std::string& str) { bool seen[26] = {false}; for (char c : str) { int index = c - 'a'; seen[index] = true; } for (bool b : seen) { if (!b) { return false; } } return true; } ``` In this function, we create the same `bool` array `seen` as before, but this time, we simply set the corresponding index to `true` for each character we encounter in the input string. After that, we iterate through the `seen` array and check if any of its elements are still `false`. If we find any, we immediately return `false`, indicating that the string does not contain all lowercase English letters. If the loop completes without finding any `false` elements, we return `true`. You can use these functions like this: ```cpp int main() { std::string str1 = abcde; std::string str2 = abcdea; std::string str3 = abcdefghijklmnopqrstuvwxyz; std::cout << std::boolalpha << hasDuplicateCharacters(str1) << std::endl; // Output: false std::cout << std::boolalpha << hasDuplicateCharacters(str2) << std::endl; // Output: true std::cout << std::boolalpha << containsAllLowercaseLetters(str3) << std::endl; // Output: true return 0; } ``` Both functions have a time complexity of O(n), where n is the length of the input string, since we are performing a constant amount of work for each character in the string. The space complexity is O(1), since we are using a fixed-size array of 26 elements.
question:Implement a queue using two stacks. A queue is a First-In-First-Out (FIFO) data structure that follows the order in which elements were added. Your Queue class needs to have the following methods: push(x): Push element x to the back of the queue. pop(): Removes the element from the front of the queue. peek(): Get the front element. empty(): Return whether the queue is empty. Example: Input: [MyQueue, push, push, peek, pop, empty] [[], [1], [2], [], [], []] Output: [null, null, null, 1, 1, false] Explanation: MyQueue myQueue = new MyQueue(); myQueue.push(1); // queue is: [1] myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue) myQueue.peek(); // return 1 myQueue.pop(); // return 1, queue is [2] myQueue.empty(); // return false
answer:Here's a C++ implementation of a queue using two stacks: ```cpp #include <stack> class MyQueue { private: std::stack<int> stackNewestOnTop; std::stack<int> stackOldestOnTop; public: MyQueue() {} void push(int x) { stackNewestOnTop.push(x); } int pop() { shiftStacks(); int front = stackOldestOnTop.top(); stackOldestOnTop.pop(); return front; } int peek() { shiftStacks(); return stackOldestOnTop.top(); } bool empty() { return stackNewestOnTop.empty() && stackOldestOnTop.empty(); } private: void shiftStacks() { if (stackOldestOnTop.empty()) { while (!stackNewestOnTop.empty()) { stackOldestOnTop.push(stackNewestOnTop.top()); stackNewestOnTop.pop(); } } } }; ``` In this implementation, we maintain two stacks: `stackNewestOnTop` and `stackOldestOnTop`. The `push` operation simply adds elements to `stackNewestOnTop`. When we need to access the front element (using `peek`) or remove it (using `pop`), we call the `shiftStacks` method. This method checks if `stackOldestOnTop` is empty, and if so, it transfers all elements from `stackNewestOnTop` to `stackOldestOnTop`, effectively reversing the order of the elements. This ensures that the front element of the queue is always at the top of `stackOldestOnTop`. The `empty` method checks if both stacks are empty. You can use the `MyQueue` class as follows: ```cpp int main() { MyQueue myQueue; myQueue.push(1); // queue is: [1] myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue) std::cout << myQueue.peek() << std::endl; // prints 1 std::cout << myQueue.pop() << std::endl; // prints 1, queue is [2] std::cout << std::boolalpha << myQueue.empty() << std::endl; // prints false return 0; } ``` This implementation has an amortized time complexity of O(1) for `push`, `pop`, `peek`, and `empty` operations, since the `shiftStacks` method is only called when necessary. The space complexity is O(n), where n is the number of elements in the queue.
question:Pasha, Masha, Tolya, and Olya ate 88 candies, with each of them eating at least one candy. Masha and Tolya together ate 57 candies, but Pasha ate the most candies. How many candies did Olya eat?
answer:: 1. Define the quantities of candies eaten by each person as follows: - Let ( P ) be the number of candies Pasha ate. - Let ( M ) be the number of candies Masha ate. - Let ( T ) be the number of candies Tolya ate. - Let ( O ) be the number of candies Olya ate. 2. According to the problem, the total number of candies eaten is: [ P + M + T + O = 88 ] 3. From the given information, we know that: [ M + T = 57 ] 4. The problem further states that Pasha ate more candies than any of the others. Hence, ( P ) is greater than both ( M ) and ( T ), and also greater than ( O ). 5. To find the minimum value of ( P ), note that ( M ) and ( T ) must each eat at least 1 candy. So, the maximum number of candies that Pasha, Masha, and Tolya could eat together would occur when one of them eats the majority. Thus if either ( M ) or ( T ) ate 28 candies, the other would eat: [ M, T geq 0 quad text{and} quad M + T = 57 ] If one of them eats 28 to maximize their total: [ text{Suppose } M = 28, quad T = 57 - 28 = 29 ] (Note that M and T are interchangeable). 6. Consequently, Pasha must have eaten: [ P > 29 quad text{(since Pasha ate more candies than Masha or Tolya)} ] 7. Hence, as Pasha ate the most candies, he must have eaten a minimum of 30 candies. Solving for ( O ): [ P + M + T + O = 88 ] Considering ( P geq 30 ) and knowing ( M + T = 57 ), the smallest ( P ) (which is 30) we get: [ P + (M + T) + O ge 30 + 57 + O = 87 + O = 88 Rightarrow O = 1 ] 8. Conclusion: [ boxed{O = 1} ]