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This approach utilizes sorting and a min-heap (or priority queue) to efficiently manage and retrieve the k-th smallest distance. The key steps are:
nums. This simplifies the process of finding the distances as all distance calculations will involve consecutive elements.Time Complexity: O(n^2 log n) due to the double loop and heap operations.
Space Complexity: O(n^2) for storing all possible distances in the heap.
1import heapq
2
3def find_kth_smallest_pair_distance(nums, k):
4 nums.sort()
5 pq = []
6 for i in range(len(nums)-1):
7 for j in range(i+1, len(nums)):
8 heapq.heappush(pq, abs(nums[j] - nums[i]))
9
10 for _ in range(k-1):
11 heapq.heappop(pq)
12 return heapq.heappop(pq)First, the array is sorted to enable orderly distance calculations. A min-heap (priority queue) is used to efficiently manage the distances and extract the k-th smallest. After computing and storing distances in the heap, we extract the smallest k elements, finally returning the k-th one.
This more efficient approach combines binary search with a two-pointer technique. The principal idea is to use binary search over distance values to pinpoint the k-th smallest distance. It involves:
nums.Time Complexity: O(n log n + n log(maxDistance))
Space Complexity: O(1) as no additional space beyond fixed variables is used.
1def find_kth_smallest_pair_distance(
This solution involves a binary search to determine the smallest possible distance. The associated helper function utilizes a two-pointer technique to count how many pairs have distances less than a proposed distance during the search process. Adjusting the binary search bounds based on this count homes in on the k-th smallest.