| 模型 | 模型答案 | 测试结果 |
|---|---|---|
| 金声玉亮2.0 | ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] heap = [] for i in range(min(k, len(nums1))): for j in range(min(k, len(nums2))): heapq.heappush(heap, (nums1[i] + nums2[j], [nums1[i], nums2[j]])) result = [] while heap and len(result) < k: _, pair = heapq.heappop(heap) result.append(pair) return result ``` [DONE] |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| deepseekr1 | To solve this problem, we need to find the k smallest pairs from two given lists, where each pair consists of one element from each list. The pairs should be sorted based on their sum, and if sums are equal, they should be sorted by the elements of the pairs. ### Approach 1. **Generate All Pairs**: Create all possible pairs from the elements of the two input lists. 2. **Heap Queue Algorithm**: Use a heap queue algorithm to efficiently find the k smallest pairs without generating and sorting all pairs explicitly. This is done using the `heapq.nsmallest` function, which allows us to specify a key for comparison. 3. **Key for Comparison**: The key for comparison is a tuple consisting of the sum of the pair, the first element of the pair, and the second element of the pair. This ensures that pairs are sorted by their sum first, and if sums are equal, they are sorted by their elements. ### Solution Code ```python import heapq def k_smallest_pairs(nums1, nums2, k): pairs = [(a, b) for a in nums1 for b in nums2] result = heapq.nsmallest(k, pairs, key=lambda pair: (pair[0] + pair[1], pair[0], pair[1])) return [list(pair) for pair in result] ``` ### Explanation 1. **Generating Pairs**: We first generate all possible pairs using list comprehensions. Each pair (a, b) is formed by taking an element `a` from the first list and an element `b` from the second list. 2. **Heapq.nsmallest**: The `heapq.nsmallest` function is used to get the k smallest pairs. The key for comparison is a lambda function that returns a tuple (sum of the pair, first element, second element). This ensures that pairs are compared first by their sum, then by the first element, and finally by the second element if necessary. 3. **Conversion to List**: The result from `heapq.nsmallest` is a list of tuples, which we convert to lists to match the expected output format. This approach efficiently leverages the heap queue algorithm to find the k smallest pairs without explicitly sorting all possible pairs, making it more efficient for larger values of k while still ensuring correctness. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| gpto1 | ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2 or k == 0: return [] m, n = len(nums1), len(nums2) heap = [] res = [] visited = set() heapq.heappush(heap, (nums1[0] + nums2[0], 0, 0)) visited.add((0, 0)) while heap and len(res) < k: sum_, i, j = heapq.heappop(heap) res.append([nums1[i], nums2[j]]) if i + 1 < m and (i + 1, j) not in visited: heapq.heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j)) visited.add((i + 1, j)) if j + 1 < n and (i, j + 1) not in visited: heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1)) visited.add((i, j + 1)) return res ``` |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| GPT-4o | To solve the task of finding k number of pairs which consist of one element from the first array and one element from the second array, we can use a min-heap to keep track of the smallest pairs. Here is the function implementation that should work as expected: ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2 or k <= 0: return [] # Min-heap to store the pairs with their sums min_heap = [] for i in range(min(len(nums1), k)): # Only need first k elements in nums1 heapq.heappush(min_heap, (nums1[i] + nums2[0], i, 0)) result = [] while min_heap and len(result) < k: sum_, i, j = heapq.heappop(min_heap) result.append([nums1[i], nums2[j]]) if j + 1 < len(nums2): heapq.heappush(min_heap, (nums1[i] + nums2[j + 1], i, j + 1)) return result # Test cases assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` This function uses a min-heap to efficiently find the k smallest pairs from the two input lists. The heap keeps track of the sum of the pairs and their respective indices in the original lists. The algorithm ensures that the pairs are generated in ascending order of their sums. This approach ensures optimal performance for generating the k smallest pairs. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| llama3.1-70B-Instruct | import heapq def k_smallest_pairs(nums1, nums2, k): queue = [] for num1 in nums1: for num2 in nums2: heapq.heappush(queue, (num1 + num2, [num1, num2])) if len(queue) > k: heapq.heappop(queue) return [pair[1] for pair in queue] |
wrong_answer 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| llama3.1-405B-Instruct | import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] heap = [] for i in range(len(nums1)): heapq.heappush(heap, (nums1[i] + nums2[0], i, 0)) result = [] while heap and k > 0: _, i, j = heapq.heappop(heap) result.append([nums1[i], nums2[j]]) if j + 1 < len(nums2): heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1)) k -= 1 return result |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| DeepSeek-Coder-V2 | ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] heap = [] for i in range(min(k, len(nums1))): for j in range(min(k, len(nums2))): heapq.heappush(heap, (nums1[i] + nums2[j], [nums1[i], nums2[j]])) result = [] while heap and len(result) < k: _, pair = heapq.heappop(heap) result.append(pair) return result ``` [DONE] This function uses a min-heap to efficiently find the k smallest pairs. It iterates through the first k elements of both arrays (or all elements if fewer than k) and pushes the sum along with the pair into the heap. Then, it pops the smallest pairs from the heap until it has k pairs or the heap is empty. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| Qwen2-72B-Instruct | 'def k_smallest_pairs(nums1, nums2, k): from itertools import product import heapq as hq pairs = list(product(nums1, nums2)) pairs = [[x, y] for x, y in pairs] smallest_pairs = hq.nsmallest(k, pairs, key=lambda x: x[0] + x[1]) return smallest_pairs' [DONE] |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| baichuan4 | To solve the task of finding `k` smallest pairs from two sorted arrays, we can use a min-heap. We will push the first `k` pairs onto the heap and then pop the smallest pairs one by one, maintaining the heap size of `k`. For each new pair, we calculate the sum of the elements and compare it with the current top of the heap. If the new sum is smaller, we pop the current top and push the new pair onto the heap. This way, after processing all possible pairs, the heap will contain the `k` smallest pairs. Here's the implementation of the `k_smallest_pairs` function: ```python import heapq def k_smallest_pairs(nums1, nums2, k): min_heap = [] # Push the first k pairs onto the heap for i in range(min(k, len(nums1))): for j in range(min(k, len(nums2))): if len(min_heap) < k: heapq.heappush(min_heap, (nums1[i] + nums2[j], i, j)) else: # If the new pair has a smaller sum than the current top, pop the top and push the new pair if nums1[i] + nums2[j] < min_heap[0][0]: heapq.heappop(min_heap) heapq.heappush(min_heap, (nums1[i] + nums2[j], i, j)) break # No need to check other pairs in nums2 # Extract the pairs from the heap result = [] while min_heap: _, i, j = heapq.heappop(min_heap) result.append([nums1[i], nums2[j]]) return result # Tests assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` This code defines the `k_smallest_pairs` function and passes the provided tests. The function uses a min-heap to keep track of the smallest pairs seen so far. It starts by pushing the first `k` pairs onto the heap and then continues to iterate through the rest of the pairs, comparing their sums with the smallest sum on the heap. If a smaller sum is found, the pair with the smallest sum is popped off the heap, and the new pair is pushed onto the heap. After processing all pairs, the function extracts the pairs from the heap and returns them as the result. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| doubao | 'from heapq import heappush, heappop def k_smallest_pairs(nums1, nums2, k): pairs = [] heap = [] for num1 in nums1: for num2 in nums2: heappush(heap, (num1 + num2, num1, num2)) for _ in range(min(k, len(heap))): _, num1, num2 = heappop(heap) pairs.append([num1, num2]) return pairs' [DONE] |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| minimax | To solve this task, we can use a min-heap to keep track of the pairs and their corresponding sums. We will generate pairs from the two arrays and push them into the heap, then pop the smallest pairs until we have k pairs. Here's the function: ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] min_heap = [] # Initialize the heap with the first pair of each element in nums1 and the first element in nums2 for i in range(min(k, len(nums1))): heapq.heappush(min_heap, (nums1[i] + nums2[0], i, 0)) result = [] # Extract the smallest pairs from the heap while k > 0 and min_heap: _, i, j = heapq.heappop(min_heap) result.append([nums1[i], nums2[j]]) k -= 1 # If there is a next element in nums2, add the new pair to the heap if j + 1 < len(nums2): heapq.heappush(min_heap, (nums1[i] + nums2[j + 1], i, j + 1)) return result # Test cases assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` This function should pass the provided test cases. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| kimi | Here is the Python function to find k number of pairs which consist of one element from the first array and one element from the second array: ```python def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] pairs = [] import heapq # Create a heap of pairs with the sum of elements and their indices heap = [(nums1[0] + nums2[0], 0, 0)] heapq.heapify(heap) while k > 0 and heap: total, i, j = heapq.heappop(heap) pairs.append([nums1[i], nums2[j]]) k -= 1 if i + 1 < len(nums1) and j < len(nums2): heapq.heappush(heap, (nums1[i+1] + nums2[j], i+1, j)) if j + 1 < len(nums2) and i < len(nums1) - 1: heapq.heappush(heap, (nums1[i] + nums2[j+1], i, j+1)) return pairs # Test cases assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` This function uses a min-heap to efficiently find the k smallest pairs. It starts by adding the pair with the smallest sum to the heap and then iteratively pops the smallest pair from the heap, adds it to the result list, and pushes the next smallest pair that can be formed with the same first element or the next element from the second array. This process continues until either the heap is empty or we have found k pairs. |
wrong_answer 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| wenxinyiyan | To solve the `k_smallest_pairs` task, we can use a heap (priority queue) to efficiently find the k smallest pairs. Here's the implementation: ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2 or k <= 0: return [] # Create a min heap to store tuples of the form (sum, num1, num2, index), # where sum is the sum of num1 and num2, and index is the index of num2 in nums2. heap = [] # Push the initial pairs onto the heap for i, num1 in enumerate(nums1): heapq.heappush(heap, (num1 + nums2[0], num1, nums2[0], 0)) # Initialize the result list result = [] # Pop k smallest pairs from the heap while heap and len(result) < k: _, num1, num2, index = heapq.heappop(heap) result.append([num1, num2]) # If there are more elements in nums2, push the next pair onto the heap if index + 1 < len(nums2): next_index = index + 1 next_num2 = nums2[next_index] heapq.heappush(heap, (num1 + next_num2, num1, next_num2, next_index)) return result ``` Now let's test the function with the provided examples: ```python assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` The function should pass all the tests. The time complexity of this solution is O(k log n), where n is the length of `nums1`. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| zhipu | To solve this problem, we can use a min-heap to efficiently find the k smallest pairs. The idea is to start by pushing the pairs formed by the first element of the first array with all elements of the second array into the heap. Then, we pop the smallest pair from the heap, and if the popped pair's first element has more elements in the first array, we push the next pair formed by the next element in the first array and the same element from the second array. Here's the implementation: ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2 or k <= 0: return [] min_heap = [] # Initialize the heap with pairs (nums1[0] + nums2[i], 0, i) for i in range(min(k, len(nums2))): heapq.heappush(min_heap, (nums1[0] + nums2[i], 0, i)) result = [] while k > 0 and min_heap: current_sum, idx1, idx2 = heapq.heappop(min_heap) result.append([nums1[idx1], nums2[idx2]]) if idx1 + 1 < len(nums1): heapq.heappush(min_heap, (nums1[idx1 + 1] + nums2[idx2], idx1 + 1, idx2)) k -= 1 return result # Test cases assert k_smallest_pairs([1,3,7], [2,4,6], 2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7], [2,4,6], 1) == [[1, 2]] assert k_smallest_pairs([1,3,7], [2,4,6], 7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` ### Explanation: 1. **Initialization**: - We start by checking if either of the input arrays is empty or if `k` is non-positive. In any of these cases, we return an empty list. - We initialize a min-heap and populate it with pairs consisting of the first element of `nums1` and each element of `nums2`. Each entry in the heap is a tuple of the form `(nums1[0] + nums2[i], 0, i)`, where `0` is the index of the first element in `nums1` and `i` is the index of the element in `nums2`. 2. **Heap Operations**: - We pop the smallest pair from the heap, add it to the result list, and then check if we can form a new pair by incrementing the index of the first element (`idx1`). If so, we push this new pair into the heap. - We continue this process until we have collected `k` pairs or the heap is empty. 3. **Return Result**: - Finally, we return the list of k smallest pairs. This approach ensures that we always process the smallest available pair, and by using a heap, we maintain efficient operations for both insertion and extraction. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| gpto1 | ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2 or k == 0: return [] m, n = len(nums1), len(nums2) heap = [] res = [] visited = set() heapq.heappush(heap, (nums1[0] + nums2[0], 0, 0)) visited.add((0, 0)) while heap and len(res) < k: sum_, i, j = heapq.heappop(heap) res.append([nums1[i], nums2[j]]) if i + 1 < m and (i + 1, j) not in visited: heapq.heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j)) visited.add((i + 1, j)) if j + 1 < n and (i, j + 1) not in visited: heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1)) visited.add((i, j + 1)) return res ``` |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |
| deepseekv3 | To solve the problem of finding the k smallest pairs from two arrays, we can use a min-heap to efficiently generate and select the pairs with the smallest sums. Here's the implementation: ```python import heapq def k_smallest_pairs(nums1, nums2, k): if not nums1 or not nums2: return [] heap = [] for num1 in nums1: for num2 in nums2: heapq.heappush(heap, (num1 + num2, [num1, num2])) result = [] for _ in range(min(k, len(heap))): result.append(heapq.heappop(heap)[1]) return result # Test cases assert k_smallest_pairs([1,3,7],[2,4,6],2) == [[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1) == [[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7) == [[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] ``` ### Explanation: 1. **Heap Initialization**: We initialize an empty heap. 2. **Pair Generation**: We iterate over each element in `nums1` and each element in `nums2`, and push the pair along with their sum into the heap. 3. **Extraction**: We then extract the k smallest pairs from the heap by popping the smallest elements from the heap. 4. **Result Construction**: The extracted pairs are stored in the `result` list and returned. This approach ensures that we efficiently find the k smallest pairs using the heap data structure, which is optimal for this kind of problem. |
pass 测试用例：assert k_smallest_pairs([1,3,7],[2,4,6],2)==[[1, 2], [1, 4]] assert k_smallest_pairs([1,3,7],[2,4,6],1)==[[1, 2]] assert k_smallest_pairs([1,3,7],[2,4,6],7)==[[1, 2], [1, 4], [3, 2], [1, 6], [3, 4], [3, 6], [7, 2]] |