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A data structure is required for storing a set of integers such that each of the following operations can be done in O (log n) time, where n is the number of elements in the set.
- Deletion of the smallest element
- Insertion of an element, if it is not already present in the set
Which of the following data structures can be used for this purpose?
- a)A heap can be used, but not a balanced binary search tree.
- b)A balanced binary search tree can be used, but not a heap.
- c)Both balanced binary search tree and heap can be used.
- d)Neither balanced binary search tree nor heap can be used.
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A data structure is required for storing a set of integers such that e...
Explanation:
To store a set of integers and perform operations like deletion of the smallest element and insertion of an element in O(log n) time, we need a data structure that can maintain an ordered collection efficiently. Two commonly used data structures that can achieve this are a heap and a balanced binary search tree.
Heap:
A heap is a complete binary tree where each node satisfies the heap property. In a min heap, each node's value is less than or equal to its children's values. Heaps are typically implemented using arrays. The smallest element can be easily accessed in O(1) time by extracting the root of the heap. Insertion of an element can be done in O(log n) time by inserting the element at the end of the heap and then performing a "heapify-up" operation to maintain the heap property.
- Deletion of the smallest element: Extracting the root of the heap takes O(1) time.
- Insertion of an element: Inserting an element at the end of the heap and performing a "heapify-up" operation takes O(log n) time.
Balanced Binary Search Tree:
A balanced binary search tree, such as AVL tree or Red-Black tree, is a binary tree where the left subtree of a node contains elements less than the node's value, and the right subtree contains elements greater than the node's value. The tree is balanced to ensure efficient operations. In a balanced binary search tree, the smallest element can be found by traversing the left subtree until reaching the leftmost leaf. Insertion of an element can be done by finding the appropriate position in the tree based on the element's value and maintaining the balance of the tree.
- Deletion of the smallest element: Traversing the left subtree until reaching the leftmost leaf takes O(log n) time.
- Insertion of an element: Finding the appropriate position in the tree and maintaining the balance takes O(log n) time.
Conclusion:
Both a heap and a balanced binary search tree can be used to store a set of integers and perform operations like deletion of the smallest element and insertion of an element in O(log n) time. However, in the given question, option 'B' is the correct answer as it states that a balanced binary search tree can be used, but not a heap. This is because a balanced binary search tree allows easy deletion of the smallest element by traversing the left subtree, while a heap requires additional operations to maintain the heap property.
To store a set of integers and perform operations like deletion of the smallest element and insertion of an element in O(log n) time, we need a data structure that can maintain an ordered collection efficiently. Two commonly used data structures that can achieve this are a heap and a balanced binary search tree.
Heap:
A heap is a complete binary tree where each node satisfies the heap property. In a min heap, each node's value is less than or equal to its children's values. Heaps are typically implemented using arrays. The smallest element can be easily accessed in O(1) time by extracting the root of the heap. Insertion of an element can be done in O(log n) time by inserting the element at the end of the heap and then performing a "heapify-up" operation to maintain the heap property.
- Deletion of the smallest element: Extracting the root of the heap takes O(1) time.
- Insertion of an element: Inserting an element at the end of the heap and performing a "heapify-up" operation takes O(log n) time.
Balanced Binary Search Tree:
A balanced binary search tree, such as AVL tree or Red-Black tree, is a binary tree where the left subtree of a node contains elements less than the node's value, and the right subtree contains elements greater than the node's value. The tree is balanced to ensure efficient operations. In a balanced binary search tree, the smallest element can be found by traversing the left subtree until reaching the leftmost leaf. Insertion of an element can be done by finding the appropriate position in the tree based on the element's value and maintaining the balance of the tree.
- Deletion of the smallest element: Traversing the left subtree until reaching the leftmost leaf takes O(log n) time.
- Insertion of an element: Finding the appropriate position in the tree and maintaining the balance takes O(log n) time.
Conclusion:
Both a heap and a balanced binary search tree can be used to store a set of integers and perform operations like deletion of the smallest element and insertion of an element in O(log n) time. However, in the given question, option 'B' is the correct answer as it states that a balanced binary search tree can be used, but not a heap. This is because a balanced binary search tree allows easy deletion of the smallest element by traversing the left subtree, while a heap requires additional operations to maintain the heap property.
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A data structure is required for storing a set of integers such that e...
A self-balancing binary search tree containing n items allows the lookup, insertion and removal of an item in O(logn) worst-case time. Since it is a BST, we can easily find out minimum element in O(nlogn).
Since heap is a balanced binary tree (or almost complete binary tree), insertion complexity for heap is O(logn).
Also, complexity to get minimum in a min heap is O(logn) because removal of root node causes a call to heapify (after removing the first element from the array) to maintain the heap tree property. But, a heap cannot be used for the above purpose as the question says – insert an element if it is not already present.
For a heap, we cannot find out if an element is present or not in O(logn).
Since heap is a balanced binary tree (or almost complete binary tree), insertion complexity for heap is O(logn).
Also, complexity to get minimum in a min heap is O(logn) because removal of root node causes a call to heapify (after removing the first element from the array) to maintain the heap tree property. But, a heap cannot be used for the above purpose as the question says – insert an element if it is not already present.
For a heap, we cannot find out if an element is present or not in O(logn).
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A data structure is required for storing a set of integers such that each of the following operations can be done in O (log n) time, where n is the number of elements in the set. Deletion of the smallest element Insertion of an element, if it is not already present in the setWhich of the following data structures can be used for this purpose?a)A heap can be used, but not a balanced binary search tree.b)A balanced binary search tree can be used, but not a heap.c)Both balanced binary search tree and heap can be used.d)Neither balanced binary search tree nor heap can be used.Correct answer is option 'B'. Can you explain this answer?
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