Dynamic Dictionaries
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Dynamic Dictionaries • Primary Operations: get(key) => search put(key, element) => insert remove(key) => delete • Additional operations: ascend() get(index) remove(index)
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Dynamic Dictionaries. Primary Operations: get(key) => search put(key, element) => insert remove(key) => delete Additional operations: ascend() get(index) remove(index). n is number of elements in dictionary. Complexity Of Dictionary Operations get(), put() and remove(). - PowerPoint PPT Presentation
Transcript of Dynamic Dictionaries
Dynamic Dictionaries
• Primary Operations: get(key) => search put(key, element) => insert remove(key) => delete
• Additional operations: ascend() get(index) remove(index)
Complexity Of Dictionary Operationsget(), put() and remove()
Data Structure Worst Case Expected
Hash Table
O(n)/O(log n) O(1)
Binary Search Tree
O(n) O(log n)
Balanced Binary Search Tree
O(log n) O(log n)
n is number of elements in dictionary
Complexity Of Other Operationsascend(), get(index), remove(index)Data Structure ascend get and
removeHash Table O(D + n log n) O(D + n log n)
Indexed BST O(n) O(n)
IndexedBalanced BST
O(n) O(log n)
D is number of buckets
The Operation put()20
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Put a pair whose key is 35.
35
The Operation remove()
Three cases:
Element is in a leaf.
Element is in a degree 1 node.
Element is in a degree 2 node.
Remove From A Leaf
Remove a leaf element. key = 7
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Remove From A Degree 1 Node
Remove from a degree 1 node. key = 40
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Remove From A Degree 1 Node (contd.)
Remove from a degree 1 node. key = 15
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Remove From A Degree 2 Node
Remove from a degree 2 node. key = 10
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Remove From A Degree 2 Node20
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Replace with largest key in left subtree (or smallest in right subtree).
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Remove From A Degree 2 Node20
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Replace with largest key in left subtree (or smallest in right subtree).
Remove From A Degree 2 Node20
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Replace with largest key in left subtree (or smallest in right subtree).
Remove From A Degree 2 Node20
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Largest key must be in a leaf or degree 1 node.
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Another Remove From A Degree 2 Node
Remove from a degree 2 node. key = 20
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Remove From A Degree 2 Node20
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Replace with largest in left subtree.
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Remove From A Degree 2 Node20
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Replace with largest in left subtree.
Remove From A Degree 2 Node18
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Replace with largest in left subtree.
Remove From A Degree 2 Node18
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Complexity is O(height).
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Yet Other Operations
• Priority Queue Motivated Operations: find max and/or min remove max and/or min initialize meld
Max And/Or Min
• Follow rightmost path to max element.• Follow leftmost path to min element.• Search and/or remove => O(h) time.
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Initialize
• Sort n elements. Initialize search tree. Output in inorder (O(n)).
• Initialize must take O(n log n) time, because it isn’t possible to sort faster than O(n log n).
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Meld10
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Meld And Merge
• Worst-case number of comparisons to merge two sorted lists of size n each is n-1.
• So, complexity of melding two binary search trees of size n each is (n).
• So, logarithmic time melding isn’t possible.
O(log n) Height Trees• Full binary trees.
Exist only when n = 2k –1.• Complete binary trees.
Exist for all n. Cannot insert/delete in O(log n) time.
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Balanced Search Trees
• Height balanced.AVL (Adelson-Velsky and Landis) trees
• Weight Balanced.• Degree Balanced.
2-3 trees2-3-4 treesred-black treesB-trees
AVL Tree
• binary tree• for every node x, define its balance factor
balance factor of x = height of left subtree of x – height of right subtree of x
• balance factor of every node x is – 1, 0, or 1
Balance Factors
0 0
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-1 0
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-1
This is an AVL tree.
Height Of An AVL Tree
The height of an AVL tree that has n nodes is at most 1.44 log2 (n+2).
The height of every n node binary tree is at least log2 (n+1).
log2 (n+1) <= height <= 1.44 log2 (n+2)
Proof Of Upper Bound On Height
• Let Nh = min # of nodes in an AVL tree whose height is h.
• N0 = 0.
• N1 = 1.
Nh, h > 1
• Both L and R are AVL trees.• The height of one is h-1.• The height of the other is h-2.• The subtree whose height is h-1 has Nh-1 nodes.• The subtree whose height is h-2 has Nh-2 nodes.• So, Nh = Nh-1 + Nh-2 + 1.
L R
Fibonacci Numbers
• F0 = 0, F1 = 1.
• Fi = Fi-1 + Fi-2 , i > 1.
• N0 = 0, N1 = 1.
• Nh = Nh-1 + Nh-2 + 1, i > 1.
• Nh = Fh+2 – 1.
• Fi ~ i/sqrt(5). sqrt
AVL Search Tree
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