1658863080

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

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")])
```

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

).

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

This work is heavily based on the implementation found here

Author: JuliaNeighbors

Source Code: https://github.com/JuliaNeighbors/BKTrees.jl

License: MIT license

1658863080

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

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")])
```

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

).

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

This work is heavily based on the implementation found here

Author: JuliaNeighbors

Source Code: https://github.com/JuliaNeighbors/BKTrees.jl

License: MIT license

1603449693

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#dentistry in keller tx #dentist in keller texas #dentist in keller tx #affordable dentist keller tx #dentist near 76244 #keller texas dental

1598227320

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

1599043260

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:**

_ D = 7 _Input:

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

_ Yes _Output:

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

_ D = 1 _Input:

```
3
/ \
3 3
\
3
```

_ No _Output:

**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:

- Perform the Post Order Traversal of the given tree and find the distance between the repeated pairs of nodes.
- Find the nodes that are repeated in the tree using unordered_map.
- For each repeated node of a particular value, find the maximum possible distance between any pair.
- If that distance is >
**D**, print “No”. - 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

1597978800

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:**

_ N = 5, S = “abcab” _Input:

_ 8 _Output:

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.

_ N = 5 S = “aabbc” _Input:

_ 5 _Output:

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
and remove it from the string.*ind* - 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