| 模型 | 模型答案 | 测试结果 |
|---|---|---|
| 金声玉亮2.0 | ```python def find_last_occurrence(arr, target): low, high = 0, len(arr) - 1 while low <= high: mid = (low + high) // 2 if arr[mid] == target and (mid == len(arr) - 1 or arr[mid + 1] != target): return mid elif arr[mid] <= target: low = mid + 1 else: high = mid - 1 return -1 ``` [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| deepseekr1 | To solve this problem, we need to find the index of the last occurrence of a given number in a sorted array. The solution should efficiently handle this using a binary search algorithm, which works well with sorted arrays and provides a time complexity of O(log n). ### Approach The approach involves using a modified binary search to locate the last occurrence of the target number. Here's the step-by-step breakdown: 1. **Initialize Pointers**: Start with two pointers, `low` set to the beginning of the array (0) and `high` set to the end of the array (length of the array minus 1). 2. **Binary Search Loop**: Continue searching while `low` is less than or equal to `high`. - Calculate the middle index (`mid`). - If the middle element is equal to the target, update the result to this index and move the `low` pointer to `mid + 1` to check for any further occurrences to the right. - If the middle element is less than the target, move the `low` pointer to `mid + 1` to search in the right half. - If the middle element is greater than the target, move the `high` pointer to `mid - 1` to search in the left half. 3. **Return Result**: After exiting the loop, the result will hold the index of the last occurrence of the target. This method ensures that we efficiently find the last occurrence by leveraging the properties of a sorted array and adjusting our search based on comparisons. ### Solution Code ```python def find_last_occurrence(arr, target): low = 0 high = len(arr) - 1 result = -1 while low <= high: mid = (low + high) // 2 if arr[mid] == target: result = mid low = mid + 1 # Look for later occurrences elif arr[mid] < target: low = mid + 1 else: high = mid - 1 return result ``` ### Explanation - **Binary Search Modification**: The key modification to the standard binary search is that when the target is found, instead of stopping, we continue searching towards the right half of the array. This helps in finding the last occurrence. - **Efficiency**: This approach maintains the O(log n) time complexity of binary search, making it efficient even for large arrays. - **Handling Edge Cases**: The algorithm handles cases where the target is not present by returning -1, though the problem constraints assume the target exists in the array. This solution efficiently and effectively finds the last occurrence of a target in a sorted array using a well-known algorithm with slight modifications to suit the problem's requirements. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| gpto1 | [BEGIN] 'def find_last_occurrence(arr, num): for idx in range(len(arr)-1, -1, -1): if arr[idx] == num: return idx return -1' [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| GPT-4o | Sure, I'll write a Python function to find the index of the last occurrence of a given number in a sorted array. Here's the function that will pass the given tests: ```python def find_last_occurrence(arr, x): low, high = 0, len(arr) - 1 result = -1 while low <= high: mid = (low + high) // 2 if arr[mid] == x: result = mid low = mid + 1 # Continue searching in the right half elif arr[mid] < x: low = mid + 1 else: high = mid - 1 return result # Tests assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 ``` This function uses a binary search algorithm to find the last occurrence of `x` in the sorted array `arr`. The variable `result` keeps track of the last seen index of `x`, and the search continues in the right half of the array whenever `x` is found. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| llama3.1-70B-Instruct | def find_last_occurrence(arr, num): for i in range(len(arr) - 1, -1, -1): if arr[i] == num: return i return -1 |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| llama3.1-405B-Instruct | def find_last_occurrence(arr, target): for i in range(len(arr) - 1, -1, -1): if arr[i] == target: return i return -1 |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| DeepSeek-Coder-V2 | ```python def find_last_occurrence(arr, target): low, high = 0, len(arr) - 1 while low <= high: mid = (low + high) // 2 if arr[mid] == target and (mid == len(arr) - 1 or arr[mid + 1] != target): return mid elif arr[mid] <= target: low = mid + 1 else: high = mid - 1 return -1 ``` [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| Qwen2-72B-Instruct | 'def find_last_occurrence(arr, num): left = 0 right = len(arr) - 1 result = -1 while left <= right: mid = left + (right - left) // 2 if arr[mid] == num: result = mid left = mid + 1 elif arr[mid] < num: left = mid + 1 else: right = mid - 1 return result' [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| baichuan4 | To find the index of the last occurrence of a given number in a sorted array, we can use a binary search approach. Here's the implementation: ```python def find_last_occurrence(arr, target): left, right = 0, len(arr) - 1 last_index = -1 while left <= right: mid = left + (right - left) // 2 if arr[mid] <= target: last_index = mid left = mid + 1 else: right = mid - 1 return last_index if arr[last_index] == target else -1 # Tests assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 ``` The `find_last_occurrence` function takes a sorted array `arr` and a target value `target`. It uses a binary search to find the last occurrence of `target` in `arr`. The search continues until the entire array is searched or the target is found. If the target is found, the function returns the index of the last occurrence of the target; otherwise, it returns -1. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| doubao | def find_last_occurrence(arr, num): for i in range(len(arr)-1, -1, -1): if arr[i] == num: return i return -1 [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| minimax | ```python def find_last_occurrence(sorted_array, target): left, right = 0, len(sorted_array) - 1 result = -1 while left <= right: mid = left + (right - left) // 2 if sorted_array[mid] == target: result = mid left = mid + 1 elif sorted_array[mid] < target: left = mid + 1 else: right = mid - 1 return result # Test cases assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 ``` This function `find_last_occurrence` uses a binary search algorithm to find the index of the last occurrence of a given number in a sorted array. The variable `result` is used to keep track of the last found index of the target value. If the middle element is equal to the target, we update `result` and continue searching to the right half of the array. If the middle element is less than the target, we search the right half; otherwise, we search the left half. The loop continues until the `left` pointer exceeds the `right` pointer. If the target is found, `result` will hold the index of the last occurrence; otherwise, it will remain `-1`. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| kimi | ```python def similar_elements(test_tup1, test_tup2): res = tuple(sorted(set(test_tup1) & set(test_tup2))) return res def is_not_prime(n): if n <= 1: return True for i in range(2, int(n**0.5) + 1): if n % i == 0: return True return False import heapq as hq def heap_queue_largest(nums, n): largest_nums = hq.nlargest(n, nums) return largest_nums def find_last_occurrence(arr, num): left, right = 0, len(arr) - 1 result = -1 while left <= right: mid = (left + right) // 2 if arr[mid] == num: result = mid left = mid + 1 elif arr[mid] < num: left = mid + 1 else: right = mid - 1 return result ``` This code provides the functions as per your tasks: 1. `similar_elements` finds the common elements in two tuples and returns them as a sorted tuple. 2. `is_not_prime` determines if a number is not prime. 3. `heap_queue_largest` finds the n largest numbers in a list using the heap queue algorithm. 4. `find_last_occurrence` finds the index of the last occurrence of a given number in a sorted array. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| wenxinyiyan | Here's the solution for each of the tasks you provided: Task 1: Finding similar elements from two tuple lists ```python def similar_elements(test_tup1, test_tup2): res = tuple(set(test_tup1) & set(test_tup2)) return res ``` This function takes two tuples as input and returns a tuple containing the common elements between them. It uses sets to find the intersection of the elements efficiently. Task 2: Identifying non-prime numbers ```python import math def is_not_prime(n): if n < 2: # Numbers less than 2 are not prime return True for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return True return False ``` This function checks whether a given number `n` is not prime. It iterates from 2 to the square root of `n` and checks if `n` is divisible by any of those numbers. If it finds a divisor, it returns `True` indicating that the number is not prime. Otherwise, it returns `False`. Note: In your original code, you had the logic reversed. I modified it to correctly identify non-prime numbers. Task 3: Finding the largest integers using the heap queue algorithm ```python import heapq as hq def heap_queue_largest(nums, n): largest_nums = hq.nlargest(n, nums) return largest_nums ``` This function uses the `heapq.nlargest()` function from the `heapq` module to find the `n` largest numbers in the given list `nums`. It returns a list containing the `n` largest numbers. Task 4: Finding the index of the last occurrence of a given number in a sorted array ```python def find_last_occurrence(arr, target): left, right = 0, len(arr) - 1 result = -1 # Initialize result to -1 (not found) while left <= right: mid = (left + right) // 2 if arr[mid] == target: result = mid # Update result to current index left = mid + 1 # Continue searching on the right side elif arr[mid] < target: left = mid + 1 else: right = mid - 1 return result ``` This function uses a modified binary search algorithm to find the index of the last occurrence of a given number `target` in a sorted array `arr`. It initializes the result to -1 (indicating not found) and then performs a binary search while updating the result whenever it finds a match. Finally, it returns the result. Note: The provided tests assume that the input array is sorted in ascending order. If the array is sorted in descending order, you would need to modify the comparison logic accordingly. |
failed 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| zhipu | To solve this problem, we can use a binary search approach to efficiently find the last occurrence of a given number in a sorted array. Here's the implementation of the function: ```python def find_last_occurrence(arr, target): left, right = 0, len(arr) - 1 result = -1 while left <= right: mid = (left + right) // 2 if arr[mid] == target: result = mid # Update result to current mid left = mid + 1 # Move left pointer to the right to find the last occurrence elif arr[mid] < target: left = mid + 1 else: right = mid - 1 return result # Test cases assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 ``` ### Explanation: 1. **Initialization**: We start by initializing `left` to 0 and `right` to the last index of the array. `result` is initialized to -1 to handle cases where the target is not found. 2. **Binary Search Loop**: - Calculate the middle index `mid`. - If `arr[mid]` equals the target, update `result` to `mid` and move the `left` pointer to `mid + 1` to continue searching in the right half for the last occurrence. - If `arr[mid]` is less than the target, move the `left` pointer to `mid + 1`. - If `arr[mid]` is greater than the target, move the `right` pointer to `mid - 1`. 3. **Return Result**: After the loop, `result` will contain the index of the last occurrence of the target, or -1 if the target is not found. This approach ensures that the function runs in O(log n) time complexity, making it efficient for large arrays. The provided test cases verify the correctness of the implementation. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| gpto1 | [BEGIN] 'def find_last_occurrence(arr, num): for idx in range(len(arr)-1, -1, -1): if arr[idx] == num: return idx return -1' [DONE] |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |
| deepseekv3 | To find the index of the last occurrence of a given number in a sorted array, you can use a binary search approach. This is efficient and works well for large datasets. Here's the function that implements this logic: ```python def find_last_occurrence(arr, target): left, right = 0, len(arr) - 1 result = -1 while left <= right: mid = left + (right - left) // 2 if arr[mid] == target: result = mid left = mid + 1 elif arr[mid] < target: left = mid + 1 else: right = mid - 1 return result ``` ### Explanation: 1. **Initialization**: Start with `left` pointing to the start of the array and `right` pointing to the end. `result` is initialized to `-1` to store the index of the last occurrence. 2. **Binary Search Loop**: - Calculate the `mid` index. - If `arr[mid]` equals the `target`, update `result` to `mid` and move `left` to `mid + 1` to search for a later occurrence. - If `arr[mid]` is less than the `target`, move `left` to `mid + 1`. - If `arr[mid]` is greater than the `target`, move `right` to `mid - 1`. 3. **Return the Result**: After the loop ends, `result` will hold the index of the last occurrence of the `target`. If the `target` is not found, `result` remains `-1`. ### Test Cases: ```python assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 ``` This function should work correctly for the provided test cases and efficiently find the last occurrence of the target in the sorted array. |
pass 测试用例：assert find_last_occurrence([2, 5, 5, 5, 6, 6, 8, 9, 9, 9], 5) == 3 assert find_last_occurrence([2, 3, 5, 8, 6, 6, 8, 9, 9, 9], 9) == 9 assert find_last_occurrence([2, 2, 1, 5, 6, 6, 6, 9, 9, 9], 6) == 6 |