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.

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.

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.

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} ]