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feat: support multiple correct answers
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,11 @@ | ||
| import type { ParsedQuestion } from './markdownParser'; | ||
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| export function isCorrectAnswer(question: ParsedQuestion, selectedIndex: number): boolean { | ||
| if (question.correctAnswer !== undefined) { | ||
| return selectedIndex + 1 === question.correctAnswer; | ||
| } | ||
| if (question.correctAnswers !== undefined) { | ||
| return question.correctAnswers.includes(selectedIndex + 1); | ||
| } | ||
| throw new Error('Question has no correct answer defined'); | ||
| } |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,141 +1,28 @@ | ||
| --- | ||
| title: The Ultimate Coding Challenge | ||
| description: Test your programming skills across various languages and concepts! | ||
| level: 2 # Questions will be sorted by level | ||
| correctAnswer: 2 | ||
| difficulty: "Expert" # Beginner, Intermediate, or Expert (optional) | ||
| title: The Integration Inquisitor | ||
| description: The Integration Inquisitor will test your knowledge of posting compositions to an OpenEHR system. | ||
| level: 2 | ||
| correctAnswers: [2, 4] | ||
| timeLimit: 5 | ||
| --- | ||
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| ## Context | ||
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| ### Introduction | ||
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| Welcome to the Ultimate Coding Challenge! This quest will test your knowledge of **programming concepts, algorithms, and problem-solving skills**. Are you ready to prove your coding prowess? | ||
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| Test of code highlighting: | ||
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| inline code: `console.log('Hello, Alice!');` | ||
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| ```python | ||
| def decorator(func): | ||
| def wrapper(*args, **kwargs): | ||
| print("Before function call") | ||
| result = func(*args, **kwargs) | ||
| print("After function call") | ||
| return result | ||
| return wrapper | ||
| ``` | ||
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| ### Question | ||
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| You're tasked with implementing a function to find the most frequent element in an array. The function should work efficiently for large arrays. Consider the following requirements: | ||
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| 1. The function should handle arrays of integers. | ||
| 2. If there are multiple elements with the same highest frequency, return any one of them. | ||
| 3. The function should have a time complexity better than O(n^2). | ||
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| ```go | ||
| func mostFrequent(arr []int) int { | ||
| maxCount := 0 | ||
| result := arr[0] | ||
| for i := range arr { | ||
| count := 0 | ||
| for j := range arr { | ||
| if arr[i] == arr[j] { | ||
| count++ | ||
| } | ||
| } | ||
| if count > maxCount { | ||
| maxCount = count | ||
| result = arr[i] | ||
| } | ||
| } | ||
| return result | ||
| } | ||
| ``` | ||
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| How would you implement this function? | ||
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| ### Outro | ||
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| __Efficient__ array manipulation and understanding of data structures are crucial skills for any programmer. This problem tests your ability to balance time complexity with code readability. | ||
| Which HTTP method and content type should you use when posting a composition to an OpenEHR system? | ||
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| ## Answers | ||
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| - Use nested loops to count occurrences of each element: | ||
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| ```python | ||
| def most_frequent(arr): | ||
| max_count = 0 | ||
| result = arr[0] | ||
| for i in range(len(arr)): | ||
| count = 0 | ||
| for j in range(len(arr)): | ||
| if arr[i] == arr[j]: | ||
| count += 1 | ||
| if count > max_count: | ||
| max_count = count | ||
| result = arr[i] | ||
| return result | ||
| ``` | ||
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| - Use a hash map to count occurrences in a single pass: | ||
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| ```python | ||
| from collections import defaultdict | ||
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| def most_frequent(arr): | ||
| count = defaultdict(int) | ||
| for num in arr: | ||
| count[num] += 1 | ||
| return max(count, key=count.get) | ||
| ``` | ||
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| - Sort the array and count consecutive elements: | ||
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| ```python | ||
| def most_frequent(arr): | ||
| arr.sort() | ||
| max_count = 1 | ||
| res = arr[0] | ||
| curr_count = 1 | ||
| for i in range(1, len(arr)): | ||
| if arr[i] == arr[i-1]: | ||
| curr_count += 1 | ||
| else: | ||
| if curr_count > max_count: | ||
| max_count = curr_count | ||
| res = arr[i-1] | ||
| curr_count = 1 | ||
| return res | ||
| ``` | ||
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| - Use a set to eliminate duplicates, then count occurrences: | ||
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| ```python | ||
| def most_frequent(arr): | ||
| return max(set(arr), key=arr.count) | ||
| ``` | ||
| - GET request with application/json content type | ||
| - POST request with application/xml content type | ||
| - PUT request with text/plain content type | ||
| - POST request with application/json content type | ||
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| ## Explanation | ||
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| The correct answer is option 2: Using a hash map to count occurrences in a single pass. | ||
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| This solution is optimal because: | ||
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| 1. It has a time complexity of O(n), where n is the length of the array. | ||
| 2. It only requires a single pass through the array. | ||
| 3. The space complexity is O(k), where k is the number of unique elements in the array. | ||
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| Here's a breakdown of how the function works: | ||
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| 1. `defaultdict(int)` creates a dictionary where any new key is automatically initialized with a value of 0. | ||
| 2. The loop `for num in arr:` iterates through each element in the array. | ||
| 3. `count[num] += 1` increments the count for each number. | ||
| 4. `max(count, key=count.get)` returns the key (number) with the highest count. | ||
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| This solution efficiently handles large arrays and meets all the specified requirements. | ||
| When posting a new composition to an OpenEHR system, you typically use a POST request with application/json or application/xml content type. This allows you to send structured data in a format that's widely supported and easy to work with. | ||
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| ## Hint | ||
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| Think about data structures that allow for fast lookups and updates. How can you keep track of counts while only passing through the array once? | ||
| Think about which HTTP method is typically used for creating new resources. |
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