Kitty's Calculations on a Tree
Kitty has a tree, T , consisting of n nodes where each node is uniquely labeled from 1 to n . Her friend Alex gave her q sets, where each set contains k distinct nodes. Kitty needs to calculate the following expression on each set: where: { u ,v } denotes an unordered pair of nodes belonging to the set. dist(u , v) denotes the number of edges on the unique (shortest) path between nodes and . Given T and q sets of k distinct nodes, calculate the expression for each set. For each set o
View Solution →Is This a Binary Search Tree?
For the purposes of this challenge, we define a binary tree to be a binary search tree with the following ordering requirements: The data value of every node in a node's left subtree is less than the data value of that node. The data value of every node in a node's right subtree is greater than the data value of that node. Given the root node of a binary tree, can you determine if it's also a binary search tree? Complete the function in your editor below, which has 1 parameter: a pointer
View Solution →Square-Ten Tree
The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the level contains subsegments of length , etc. In other words, every level (for every ) of square-ten
View Solution →Set .intersection() Operation
.intersection() The .intersection() operator returns the intersection of a set and the set of elements in an iterable. Sometimes, the & operator is used in place of the .intersection() operator, but it only operates on the set of elements in set. The set is immutable to the .intersection() operation (or & operation). >>> s = set("Hacker") >>> print s.intersection("Rank") set(['a', 'k']) >>> print s.intersection(set(['R', 'a', 'n', 'k'])) set(['a', 'k']) >>> print s.intersection(['
View Solution →Set .union() Operation
.union() The .union() operator returns the union of a set and the set of elements in an iterable. Sometimes, the | operator is used in place of .union() operator, but it operates only on the set of elements in set. Set is immutable to the .union() operation (or | operation). Example >>> s = set("Hacker") >>> print s.union("Rank") set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n']) >>> print s.union(set(['R', 'a', 'n', 'k'])) set(['a', 'R', 'c', 'r', 'e', 'H', 'k', 'n']) >>> print s
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