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Practice Final
... A police department wants to maintain a database of up to 1800 license-plate numbers of people who receive frequent tickets so that it can be determined very quickly whether or not a given license plate is in the database. Speed of response is very important; efficient use of memory is also importan ...
... A police department wants to maintain a database of up to 1800 license-plate numbers of people who receive frequent tickets so that it can be determined very quickly whether or not a given license plate is in the database. Speed of response is very important; efficient use of memory is also importan ...
One-dimensional range searching. Two-dimensional range
... µ0 could be spliced out of the tree. A similar treatement for µ00 , but this time you splice out nodes greater than µ00 . Then the answer can be obtained by a pre-order walk of the tree, enumerating the nodes as you go, will cost in time complexity O(k). The total of all three operations being O(2 l ...
... µ0 could be spliced out of the tree. A similar treatement for µ00 , but this time you splice out nodes greater than µ00 . Then the answer can be obtained by a pre-order walk of the tree, enumerating the nodes as you go, will cost in time complexity O(k). The total of all three operations being O(2 l ...
18-BinaryTrees
... root an external reference to the root node (or null if empty) leaf a node that has no children branch any internal node; neither the root node nor a leaf parent a node that refers to this one root child a node that this node refers to ...
... root an external reference to the root node (or null if empty) leaf a node that has no children branch any internal node; neither the root node nor a leaf parent a node that refers to this one root child a node that this node refers to ...
Selection sort
... the smallest element of L" could take many comparisons. Need to look more carefully at this part of the loop. (The other part, adding an element to X, also depends on how we store X, but can be done in constant time for most reasonable implementations and in any case doesn't require any comparisons, ...
... the smallest element of L" could take many comparisons. Need to look more carefully at this part of the loop. (The other part, adding an element to X, also depends on how we store X, but can be done in constant time for most reasonable implementations and in any case doesn't require any comparisons, ...
Binary Trees - CIS @ UPenn
... • A binary tree is defined recursively: it consists of a root, a left subtree, and a right subtree • To traverse (or walk) the binary tree is to visit each node in the binary tree exactly once • Tree traversals are naturally recursive • Since a binary tree has three “parts,” there are six possible w ...
... • A binary tree is defined recursively: it consists of a root, a left subtree, and a right subtree • To traverse (or walk) the binary tree is to visit each node in the binary tree exactly once • Tree traversals are naturally recursive • Since a binary tree has three “parts,” there are six possible w ...
DataStructuresExamSupp12 - School of Computer Science
... Understand how to build binary trees, heaps, graphs together with their internal representations and relevant algorithms. Appreciate differences between basic complexity classes of algorithms (constant, linear, quadratic and logarithmic). ...
... Understand how to build binary trees, heaps, graphs together with their internal representations and relevant algorithms. Appreciate differences between basic complexity classes of algorithms (constant, linear, quadratic and logarithmic). ...
Binary Search Trees A Generic Tree Binary Trees
... Worst Case : Could be n nodes from root to the bottom. Searches on BST : On average require about 2 log n comparisons on a tree with n nodes. Proof: # of compares = 1 + distance of node to the root Adding over all nodes gives internal path length If C n = average internal path length of BST with n n ...
... Worst Case : Could be n nodes from root to the bottom. Searches on BST : On average require about 2 log n comparisons on a tree with n nodes. Proof: # of compares = 1 + distance of node to the root Adding over all nodes gives internal path length If C n = average internal path length of BST with n n ...
Lecture 8 1 Overview 2 Motivation for Binary Search Trees
... we delete a key from a node with minimum keys (1 for a (2,3) tree), we merge the node with its neighbor to create a node with more keys. We can then delete the key from this new node. For the case of (2,3) trees, the amortized cost of inserting or deleting a key is O(log n). To illustrate why our am ...
... we delete a key from a node with minimum keys (1 for a (2,3) tree), we merge the node with its neighbor to create a node with more keys. We can then delete the key from this new node. For the case of (2,3) trees, the amortized cost of inserting or deleting a key is O(log n). To illustrate why our am ...
Chapter 8 - CENG METU
... – Components can be stored one after the other in a contiguous block – Components can be stored in separate locations identified by pointers ...
... – Components can be stored one after the other in a contiguous block – Components can be stored in separate locations identified by pointers ...
1 Review: Data Structures 2 Heaps: A quick review
... is a binary tree which stores values at nodes with the following order relation. The value at a node is larger than the value of its left child and is at most the value of its right child. Exercise. Write a procedure to output the values of a binary search tree in sorted order. It should visit each ...
... is a binary tree which stores values at nodes with the following order relation. The value at a node is larger than the value of its left child and is at most the value of its right child. Exercise. Write a procedure to output the values of a binary search tree in sorted order. It should visit each ...
Persistent Data Structures 2.1 Introduction and motivation
... How about time bounds? Well, access time gets an O(1) slowdown (plus an additive O(log m) cost for finding the correct root), just as we’d hoped! (We must check the modification box on each node we access, but that’s it.) Time and space for modifications require amortized analysis. A modification takes ...
... How about time bounds? Well, access time gets an O(1) slowdown (plus an additive O(log m) cost for finding the correct root), just as we’d hoped! (We must check the modification box on each node we access, but that’s it.) Time and space for modifications require amortized analysis. A modification takes ...
data structuer Lecture 1
... • An array is a list of finite number n of similar data elements referenced respectively by a set of n consecutive numbers . • Arrays can have n dimensions • One-dimentional arrays are arrays in which each element is referenced by one subscript. • A two- dimentional array is a collection of similar ...
... • An array is a list of finite number n of similar data elements referenced respectively by a set of n consecutive numbers . • Arrays can have n dimensions • One-dimentional arrays are arrays in which each element is referenced by one subscript. • A two- dimentional array is a collection of similar ...
Solution - University of Toronto
... Why do they work? We analyze the first of the two methods (the other is proved similarly). If the condition in line 2 holds then we can avoid looking at the left subtree as all elelemtns there are clearly too small. Otherwise, all elements of the right subtree must be counted, the root must be count ...
... Why do they work? We analyze the first of the two methods (the other is proved similarly). If the condition in line 2 holds then we can avoid looking at the left subtree as all elelemtns there are clearly too small. Otherwise, all elements of the right subtree must be counted, the root must be count ...
Lecture 4
... pointers, and whether the tree is a full binary tree. From the Full Binary Tree Theorem: Half of the pointers are null. Ex: Full tree, all nodes store data, with two pointers to children Total space required is (2p + d)n (a tree of n nodes, p: space of a ...
... pointers, and whether the tree is a full binary tree. From the Full Binary Tree Theorem: Half of the pointers are null. Ex: Full tree, all nodes store data, with two pointers to children Total space required is (2p + d)n (a tree of n nodes, p: space of a ...