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Start learning 50% faster. Sign in nowA min-heap is a complete binary tree where each node is smaller than or equal to its children. The smallest element is always at the root of the heap. Removing the smallest element involves two main steps: 1. Replace the root with the last element: This is a O(1) operation, as it simply involves accessing and swapping nodes. 2. Restore the heap property using the heapify operation: The heapify process involves comparing the new root with its children and swapping it with the smaller child until the heap property is restored. This operation depends on the height of the tree. Since a heap is a complete binary tree, its height is O(logn), where nnn is the number of nodes. The heapify operation traverses from the root to the lowest level, making the overall complexity O(logn). Option 2, O(logn), is correct because restoring the heap property after removal involves percolating the new root down the tree, which is logarithmic concerning the number of elements. Why Other Options Are Incorrect? 1. O(1): This is incorrect because although removing the root itself is O(1), the heapify process involves multiple comparisons and swaps, which make the operation logarithmic. 2. O(n): Restoring the heap property requires traversing the height of the tree, not all n elements. O(n) complexity would apply to operations like building a heap, not individual removals. 3. O(nlogn): This complexity applies to heap sort, where we repeatedly remove elements and restore the heap property. Here, only a single removal is being considered. 4. O(n2): This is an unreasonable complexity for any heap operation, as heaps are designed for efficiency. Thus, O(logn) is the correct and expected complexity for this operation.
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