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By using a Min-Heap, we can efficiently keep track of the k largest elements. Maintain a heap of size k, and for each element in the array, decide whether to add it to the heap. If you add an element to the heap when it is full, remove the minimum element to maintain the size of the heap as k. At the end, the root of the heap will be the kth largest element.
Time Complexity: O(n log n), where n is the size of the array, because of the sorting operation.
Space Complexity: O(1), as we are modifying the input array in-place.
1import heapq
2
3def findKthLargest(nums, k):
4 minHeap = []
5 for num in nums:
6 heapq.heappush(minHeap, num)
7 if len(minHeap) > k:
8 heapq.heappop(minHeap)
9 return minHeap[0]
10
11nums = [3,2,1,5,6,4]
12k = 2
13print(findKthLargest(nums, k))We use Python's heapq module to maintain a min-heap. For each element in the array, the element is pushed to the heap. If the heap size exceeds k, the smallest element is popped. The smallest element remaining in the heap is the kth largest element.
Quickselect is an efficient selection algorithm to find the kth smallest element in an unordered list. The algorithm is related to the quicksort sorting algorithm. It works by partitioning the data, similar to the partitioning step of quicksort, and is based on the idea of reducing the problem size through partitioning.
Average-case Time Complexity: O(n), but worst-case O(n²).
Space Complexity: O(1) for iterative solution, O(log n) for recursive calls due to stack depth.
1import
Randomized quickselect is utilized for the Java interpretation. A random pivot ensures a lower likelihood of the worst-case performance. Swapping and managing elements relative to pivot helps finalize the kth position.