Monty  Boehm

Monty Boehm

1658863080

BKTrees.jl: Burkhard-Keller Trees Implementation

BKTrees.jl

Julia implementation of BKTrees based on the Python implementation found here.

Documentation

This short example illustrates the usage of the B-K trees for approximate string matching:

julia> using Pkg
       Pkg.activate(".")
       using BKTrees
       using Random
       using StringDistances
       
       lev(x,y) = evaluate(Levenshtein(), x, y)
       dictionary = [randstring(10) for _ in 1:100_000]  # random dictionary
       bkt = BKTree(lev, dictionary)  # build tree
       
       target = randstring(10)  # target string
       
       # search for best 3 matches with distance < 10
       found = find(bkt, target, 10, k=3)
       @show target, found

#(target, found) = ("RIqWKU2A38", Tuple{Float64,String}[(7.0, "uIfPK02wH9"), (7.0, "RIqTF8YC6O"), (7.0, "XMqWKG1GHN")])

Installation

The installation can be done through the usual channels (manually by cloning the repository or installing it though the julia REPL).

Introductio

For more information on BK-trees check out https://en.wikipedia.org/wiki/BK-tree and https://github.com/Jetsetter/pybktree.

Acknowledgments

This work is heavily based on the implementation found here

Author: JuliaNeighbors
Source Code: https://github.com/JuliaNeighbors/BKTrees.jl 
License: MIT license

#julia #tree #string 

What is GEEK

Buddha Community

BKTrees.jl: Burkhard-Keller Trees Implementation
Monty  Boehm

Monty Boehm

1658863080

BKTrees.jl: Burkhard-Keller Trees Implementation

BKTrees.jl

Julia implementation of BKTrees based on the Python implementation found here.

Documentation

This short example illustrates the usage of the B-K trees for approximate string matching:

julia> using Pkg
       Pkg.activate(".")
       using BKTrees
       using Random
       using StringDistances
       
       lev(x,y) = evaluate(Levenshtein(), x, y)
       dictionary = [randstring(10) for _ in 1:100_000]  # random dictionary
       bkt = BKTree(lev, dictionary)  # build tree
       
       target = randstring(10)  # target string
       
       # search for best 3 matches with distance < 10
       found = find(bkt, target, 10, k=3)
       @show target, found

#(target, found) = ("RIqWKU2A38", Tuple{Float64,String}[(7.0, "uIfPK02wH9"), (7.0, "RIqTF8YC6O"), (7.0, "XMqWKG1GHN")])

Installation

The installation can be done through the usual channels (manually by cloning the repository or installing it though the julia REPL).

Introductio

For more information on BK-trees check out https://en.wikipedia.org/wiki/BK-tree and https://github.com/Jetsetter/pybktree.

Acknowledgments

This work is heavily based on the implementation found here

Author: JuliaNeighbors
Source Code: https://github.com/JuliaNeighbors/BKTrees.jl 
License: MIT license

#julia #tree #string 

Steve Morkel

1603449693

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Remove all leaf nodes from a Generic Tree or N-ary Tree

Given a Generic tree, the task is to delete the leaf nodes from the tree.

** Examples:**

Input: 
              5
          /  /  \  \
         1   2   3   8
        /   / \   \
       15  4   5   6 

Output:  
5 : 1 2 3
1 :
2 :
3 :

Explanation: 
Deleted leafs are:
8, 15, 4, 5, 6

Input:      
              8
         /    |    \
       9      7       2
     / | \    |    / / | \ \
    4  5 6    10  11 1 2  2 3
Output:  
8: 9 7 2
9:
7:
2:

**Approach: **Follow the steps given below to solve the problem

  • Take tree into the vector.
  • Traverse the tree and check the condition:
  • If current node is leaf then
  • Delete the leaf from vector
  • Else
  • Recursively call for every child.

Below is the implementation of the above approach:

#data structures #recursion #tree #n-ary-tree #tree-traversal #data analysis

Check if all the Nodes in a Binary Tree having common values are at least D distance apart

Given a Binary Tree and an integer D, the task is to check if the distance between all pairs of same node values in the Tree is ? D or not. If found to be true, then print Yes. Otherwise, print No.

Examples:

Input:_ D = 7 _

                1
              /   \ 
             2     3
            / \   /  \ 
           4   3  4   4

Output:_ Yes _

Explanation:

_The repeated value of nodes are 3 and 4. _

_The distance between the two nodes valued 3, is 3. _

_The maximum distance between any pair of nodes valued 4 is 4. _

Therefore, none of the distances exceed 7

Input:_ D = 1 _

          3
         / \
        3   3
             \
              3

Output:_ No _

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

**Approach: **

The idea is to observe that the problem is similar to finding the distance between two nodes of a tree. But there can be multiple pairs of nodes for which we have to find the distance. Follow the steps below:

  1. Perform the Post Order Traversal of the given tree and find the distance between the repeated pairs of nodes.
  2. Find the nodes that are repeated in the tree using unordered_map.
  3. For each repeated node of a particular value, find the maximum possible distance between any pair.
  4. If that distance is > D, print “No”.
  5. If no such node value is found having a pair containing that value, exceeding **D, **then print “Yes”.

#greedy #recursion #searching #tree #binary tree #frequency-counting #postorder traversal #tree-traversal

Mikel  Okuneva

Mikel Okuneva

1597978800

Find value after N operations to remove N characters of string S with given constraints

Given a string** S** of Size N. Initially, the value of count is 0. The task is to find the value of count after N operations to remove all the N characters of the given string S where each operation is:

  • In each operation, select an alphabetically the smallest character in the string S and remove that character from S and add its index to count.
  • If multiple characters are present then remove the character having the smallest index.

Note: Consider string as 1-based indexing.

Examples:

Input:_ N = 5, S = “abcab” _

Output:_ 8 _

Explanation:

_Remove character ‘a’ then string becomes “bcab” and the count = 1. _

_Remove character ‘a’ then string becomes “bcb” and the count = 1 + 3 = 4. _

_Remove character ‘b’ then string becomes “cb” and the count = 4 + 1 = 5. _

_Remove character ‘b’ then string becomes “c” and the count = 5 + 2 = 7. _

Remove character ‘c’ then string becomes “” and the count = 7 + 1 = 8.

Input:_ N = 5 S = “aabbc” _

Output:_ 5 _

Explanation:

The value after 5 operations to remove all the 5 characters of String aabbc is 5.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

**Naive Approach: **The idea is to check if the string is empty or not. If it is not empty then following are the steps to solve the problem:

  • Find the first occurrence of the smallest alphabets in the current string, let’s say ind and remove it from the string.
  • Increase the count by **ind **+ 1.
  • Repeat the above step until the string becomes empty.

Below is the implementation of the above approach:

  • C++
  • Java
  • C#

// C++ program for the above approach

#include <iostream>

#include <string>

**using** **namespace** std;

// Function to find the value after

// N operations to remove all the N

// characters of string S

**void** charactersCount(string str, **int** n)

{

**int** count = 0;

// Iterate till N

**while** (n > 0) {

**char** cur = str[0];

**int** ind = 0;

**for** (``**int** j = 1; j < n; j++) {

**if** (str[j] < cur) {

cur = str[j];

ind = j;

}

}

// Remove character at ind and

// decrease n(size of string)

str.erase(str.begin() + ind);

n--;

// Increase count by ind+1

count += ind + 1;

}

cout << count << endl;

}

// Driver Code

**int** main()

{

// Given string str

string str = "aabbc"``;

**int** n = 5;

// Function call

charactersCount(str, n);

**return** 0;

}

Output:

5

Time Complexity:_ O(N2)_

Auxiliary Space:_ O(1)_

**Efficient Approach: **This problem can be solved using the concept of Binary Index Tree or Fenwick Tree. Below are the steps:

  • Initially, Store the values of indices of all the characters/alphabet in increasing order.
  • Start with the smallest alphabet ‘a’ and remove all ‘a’s in the order of there occurrence. After removing, select the next alphabet ‘b’, and repeat this step until all alphabets are removed.
  • Removing the character means that its corresponding value in the array is changed to 0, indicating that it is removed.
  • Before removing, find the position of the character in the string using the sum() method in Fenwick Tree and add the position value to the count.
  • After removing all the characters in the string, the value of count is obtained.

#advanced data structure #divide and conquer #strings #tree #segment-tree #trees