get id value from url

import numpy as np

def pagerank(M, num_iterations=100, d=0.85):
    N = M.shape[1]
    v = np.random.rand(N, 1)
    v = v / np.linalg.norm(v, 1)
    iteration = 0
    while iteration < num_iterations:
        iteration += 1
        v = d * np.matmul(M, v) + (1 - d) / N
    return v

import java.util.Stack;

/**
 * Java Program to implement a binary search tree. A binary search tree is a
 * sorted binary tree, where value of a node is greater than or equal to its
 * left the child and less than or equal to its right child.
 * 
 * @author WINDOWS 8
 *
 */
public class BST {

    private static class Node {
        private int data;
        private Node left, right;

        public Node(int value) {
            data = value;
            left = right = null;
        }
    }

    private Node root;

    public BST() {
        root = null;
    }

    public Node getRoot() {
        return root;
    }

    /**
     * Java function to check if binary tree is empty or not
     * Time Complexity of this solution is constant O(1) for
     * best, average and worst case. 
     * 
     * @return true if binary search tree is empty
     */
    public boolean isEmpty() {
        return null == root;
    }

    
    /**
     * Java function to return number of nodes in this binary search tree.
     * Time complexity of this method is O(n)
     * @return size of this binary search tree
     */
    public int size() {
        Node current = root;
        int size = 0;
        Stack<Node> stack = new Stack<Node>();
        while (!stack.isEmpty() || current != null) {
            if (current != null) {
                stack.push(current);
                current = current.left;
            } else {
                size++;
                current = stack.pop();
                current = current.right;
            }
        }
        return size;
    }


    /**
     * Java function to clear the binary search tree.
     * Time complexity of this method is O(1)
     */
    public void clear() {
        root = null;
    }

}

 [[pdoResources?
                                        &parents=`0`
                                        &resources=`[[++company]], [[++new-products]], [[++used-products]], [[++koptika]], [[++services]], [[++brands]], [[++news]], [[++contact]]`
                                        &limit=`20`
                                        &depth=`10`
                                        &tpl=`footerLinksTPL`
                                        &sortby=`FIELD(modResource.id, [[++company]], [[++new-products]], [[++used-products]], [[++koptika]], [[++services]], [[++brands]], [[++news]], [[++contact]] )`
                                        &sortdir=`ASC`
                                        &showHidden=`1`
                                        &context=`[[!getContext]]`
                                    ]]

optimizely.get('visitor').visitorId

console.table(optimizely.get('state').getVariationMap() );

 const searchParams = new URLSearchParams(window.location.search);
  const id = searchParams.get('id');
  console.log(id)

​//Search a value with incorrect spaces.. 

var input = 'gu ara nteed rate';
var table = 'core_company';
var field = 'name';

var searchName = input.replace(/\s/g, '').toLowerCase();
var gr = new GlideRecord(table);
gr.addEncodedQuery(field + 'ISNOTEMPTY');
gr.query();
while (gr.next()) {
  	var fieldString = gr.getValue(field).replace(/\s/g, '').toLowerCase();
 
    if (fieldString.indexOf(searchName) != -1)
        gs.info('FOUND: "' + gr[field].getDisplayValue() + '"');
}
/*
Example:

2021-03-15T18:05:40.674Z: FOUND: "Guaranteed Rate Insurance"
2021-03-15T18:05:40.677Z: FOUND: "Guaranteed Rate, Inc2"
2021-03-15T18:05:40.678Z: FOUND: "Guaranteed Rate, Inc."
2021-03-15T18:05:40.680Z: FOUND: "Guaranteed Rate, Inc"
2021-03-15T18:05:40.682Z: FOUND: "Guaranteed Rate Affinity"

*/

// Efficient Code : Time Complexity : O(R + C)
 
import java.util.*;
import java.io.*;
 
class GFG 
{ 
	static int R = 4, C = 4;

	static void search(int mat[][], int x)
	{
		int i  = 0, j = C - 1;

		while(i < R && j >= 0)
		{
			if(mat[i][j] == x)
			{
				System.out.println("Found at (" + i + ", " + j + ")");
				return;
			}
			else if(mat[i][j] > x)
			{
				j--;
			}
			else
			{
				i++;
			}
		}
		System.out.println("Not Found");
	}

	public static void main(String args[]) 
    {
        int arr[][] = {{10, 20, 30, 40},
    				   {15, 25, 35, 45},
    				   {27, 29, 35, 45},
    				   {32, 33, 39, 50}};
    	int x = 29;	   

    	search(arr, x);

		
    } 
}
 
 
 
 
// Naive Code : Time Complexity : O(R * C)
 
	static int R = 4, C = 4;

	static void search(int mat[][], int x)
	{
		for(int i = 0; i < R; i++)
		{
			for(int j = 0; j < C; j++)
			{
				if(mat[i][j] == x)
				{
					System.out.println("Found at (" + i + ", " + j + ")");
					
					return;
				}
			}
		}

		System.out.println("Not Found");
	}

// Search Highlighting
function buson_highlight_search_results($text)
{
   if (is_search()) {
      $pattern = '/('. join('|', explode(' ', get_search_query())).')/i';
      $text = preg_replace($pattern, '<span class="search-highlight">\0</span>', $text);
   }
   return $text;
}
add_filter('the_content', 'buson_highlight_search_results');
add_filter('the_excerpt', 'buson_highlight_search_results');
add_filter('the_title', 'buson_highlight_search_results');

function philosophy_search_form($form)
{
   $home_dir = home_url('/');
   $label = __('Search for:', 'philosophy');
   $btn_label = __('Search', 'philosophy');
   $search_form = <<<FORM
   <form role="search" method="get" class="header__search-form" action="{$home_dir}">
      <label>
            <span class="hide-content">{$label}</span>
            <input type="search" class="search-field" placeholder="Type Keywords" value="" name="s" title="{$label}" autocomplete="off">
      </label>
      <input type="hidden" name="post_type" value="book">
      <input type="submit" class="search-submit" value="{$btn_label}">
   </form>
   FORM;

   return $search_form;
}
add_filter('get_search_form', 'philosophy_search_form');

haystack.indexOf(haystack.find(arr => arr.includes(search)));
<- 2

# //Lists are linear but trees are nonlinear 
# //Often working with binary trees and specifically binary search trees 
 
# //SL is techically a special case tree 
 
# //Root: top node of tree
# //Child: node directly connected to another node when moving away from root 
# //Parent: converse notion of child 
# //Siblings: group of node with same parent 
# //Leaf: node with no children 
# //Edge: connection betweeen one node and another (often arrows) 
 
# //Used for HTML DOM, network routing, abstract syntax trees, folder organization, AI, best possible move (decision tree)
 
# //We will focus on binary search tree (can only have at most 2 child nodes)
# //For each node, everything to left is smaller. Right >> bigger
 


class Node: 
  def __init__(self, value): 
    self.value = value 
    self.left = None 
    self.right = None 

class BinarySearchTree: 
  def __init__(self): 
    self.root = None 

  #adds number to correct place 
  def insert(self, value): 
    #creates new Node   
    new_node = Node(value) 
    #start at root 
    #if no root exists, root becomes new_node 
    if self.root == None: 
      self.root = new_node 
      return self 
    
    current = self.root 
    
    while True: 
      #to handle special case where value is same as current node 
      #you can return None or you can add a counter property 
      if value == current.value: 
        return None 
      #check to see if value is less than current 
      if value < current.value: 
        #check to see if there is node to left 
        #if not, add new_node to left 
        if current.left == None: 
          current.left = new_node 
          return self
        #if there is node to left, move to that node and repeat 
        current = current.left 
      #check to see if value is greater than current 
      else: 
        #check to see if there is node to right 
        #if not, add new_node to right 
        if current.right == None: 
          current.right = new_node 
          return self 
        #if there is node to right, move to that node and repeat 
        current = current.right 
  
  #find() returns value while contains() returns boolean 
  #find will point to entire node 
  def find(self, value): 
    #check to see if there is root and if not, we are done! 
    if self.root == None: 
      return False 
    current = self.root 
    found = False 
    
    while current and not found: 
      #if value is less than current 
      if value < current.value: 
        #check to see if there is node to left 
        #if there is, move to that node and repeat 
        current = current.left 
        #if not, we ar edone because current will no longer exist 
      #if value is greater than current 
      elif value > current.value: 
        #check to see if there is node to right 
        #if there is, move to that node and repeat 
        current = current.right 
        #if not, we are done because current will no longer exist 
      else: 
        found = True 
    
    if not found:
      return None 
    return current 
    
  def contains(self, value): 
    if self.root == None: 
      return False 
    current = self.root 
    found = False 
    
    while current and not found: 
      if value < current.value: 
        current = current.left 
      elif value > current.value: 
        current = current.right 
      else: 
        return True 
    return False 
          
t = BinarySearchTree()
t.insert(3)
t.insert(5)
t.insert(2)
t.insert(-1)
t.insert(6)
t.find(6) 
t.contains(6)

# // Insertion and Search: O(log N) VERY GOOD 
# // Not guaranteed if BST is a one sided tree (choose a different root) 
 
 
 
# //      10
# //   5     13
# // 2  7  11  16
 
# // tree = BinarySearchTree()
# // tree.insert(10)
# // tree.insert(5)
# // tree.insert(13)
# // tree.insert(11)
# // tree.insert(2)
# // tree.insert(16)
# // tree.insert(7)
# // tree.find(7) >> will point to node with 7  
# // tree.contains(6) >> False 
 
 
 
 
 
# //Primitive way of making Tree before insert() 
# // tree = BinarySearchTree()
# // tree.root = Node(10)
# // tree.root.right = Node(15)
# // tree.root.left = Node(7)
# // tree.root.left.right = Node(9)

# //Tree Traversal (visit every node once) 
# //Heavy on recursion 
 
# //Breadth-first (BFS) vs. Depth-first (DFS) 
 
# // DFS: 
# // 1) in order (go in order of value)
# // 2) pre order (start at root)
# // 3) post order (start at botom)
 
 
# //When to use BFS vs. DFS: 
# //For a fully loaded (wide) tree, BFS queue will be overloaded in the start (overloaded space complexity)
# //For a longer tree, DFS will take more memory (more rare, Trees are usually wide)
# //Big O for both is same 
 
# //In order: useful if you want sorted array at end (you don't really know what the root is bc it's in the middle)
# //PreOrder: useful to "export a tree" so that it can be copied bc it flattens it 

class Node: 
  def __init__(self, value): 
    self.value = value 
    self.left = None 
    self.right = None 
 
class BinarySearchTree: 
  def __init__(self): 
    self.root = None 
 
  #adds number to correct place 
  def insert(self, value): 
    #creates new Node   
    new_node = Node(value) 
    #start at root 
    #if no root exists, root becomes new_node 
    if self.root == None: 
      self.root = new_node 
      return self 
    
    current = self.root 
    
    while True: 
      #to handle special case where value is same as current node 
      #you can return None or you can add a counter property 
      if value == current.value: 
        return None 
      #check to see if value is less than current 
      if value < current.value: 
        #check to see if there is node to left 
        #if not, add new_node to left 
        if current.left == None: 
          current.left = new_node 
          return self
        #if there is node to left, move to that node and repeat 
        current = current.left 
      #check to see if value is greater than current 
      else: 
        #check to see if there is node to right 
        #if not, add new_node to right 
        if current.right == None: 
          current.right = new_node 
          return self 
        #if there is node to right, move to that node and repeat 
        current = current.right 
        
  #Breadth first search iterative using queue 
  def BFS(self):
    node = self.root 
    #we will return data 
    data = [] 
    queue = [] 
    #place root node inside queue (recall FIFO)
    queue.append(node) 
    #while queue is not empty 
    while len(queue) > 0: 
      #dequeue node from queue and append to data  
      node = queue.pop(0)
      data.append(node) 
      #if there is left on node dequeued, add to queue 
      if node.left: 
        queue.append(node.left) 
      #if there is right on node dequeued, add to queue 
      if node.right: 
        queue.append(node.right) 
      #above two lines of code could be changed if it was a ternary tree, etc. instead of binary 
      #just loop for all children 
    return data 
  
  #parent down uses recursive 
  def DFSPreoder(self): 
    data = [] 
    #if you want to DFS not from root, create a variable here named current and specify which node to start from 
    #helper function 
    def traverse(node): 
      #all of root node's left will happen first, then right 
      #for other types of DFS, just tweak this order 
      data.append(node.value) 
      if node.left:
        traverse(node.left)
      if node.right: 
        traverse(node.right)
    
    traverse(self.root) 
    return data 
  
  #children up, root should be last value 
  def DFSPostOrder(self): 
    data = []
    
    def traverse(node): 
      if node.left:
        traverse(node.left)
      if node.right: 
        traverse(node.right)
      data.append(node.value)
    
    traverse(self.root)
    return data 
  
  #result data list should be sorted 
  def DFSInOrder(self):
    data = []
    def traverse(node): 
      if node.left: 
        traverse(node.left)
      data.push(node.value)
      if node.right: 
        traverse(node.right)
    traverse(self.root)
    return data 

t = BinarySearchTree() 
t.insert(1)
t.insert(5)
t.insert(6)
t.insert(2)
t.insert(0) 
t.insert(-1)
t.insert(7) 

print(t.DFSInOrder())

#
# Binary trees are already defined with this interface:
# class Tree(object):
#   def __init__(self, x):
#     self.value = x
#     self.left = None
#     self.right = None
import math 
def solution(t):
    if t is None: return [] 
    
    stack = [t] 
    result = []
    
    while len(stack) > 0: 
        result.append(max(tree.value for tree in stack)) 
        next_row = [tree.left for tree in stack if tree.left] + [tree.right for tree in stack if tree.right]
        stack = next_row 
    return result 


 #1. Add max value of ‘stack’ to result. 2. Create a new list of all next values for each value in stack. 3. redefine stack to this newly made list. 4. repeat 


#alternate solution 
def solution(t):
    largestValues = []
    q = []
    height = 0
    if t:
        q.append([t, height])
        while q:
            item = q.pop()
            node = item[0]
            currentHeight = item[1]
            if node.left:
                q.insert(0, [node.left, currentHeight + 1] )
            if node.right:
                q.insert(0, [node.right, currentHeight + 1])
            checkLargest(node.value, currentHeight, largestValues)
            
    return largestValues
    
        
def checkLargest(value, height, largestValues):
    if height == len(largestValues):
        largestValues.append(value)
    else:
        if largestValues[height] < value:
            largestValues[height] = value

class Tree(object):
  def __init__(self, x, left=None, right=None):
    self.value = x
    self.left = left
    self.right = right 

x = Tree(1, Tree(0, Tree(1), Tree(3)), Tree(4))


def solution(t): 
    if not t: 
        return 0 
    
    stack = [(t, 0)]
    branchesSum = 0 
    
    while stack: 
        node, v = stack.pop() 
        if node.left or node.right:
        #depth first search 
        # the v is a little confusing to understand but easier to see in python tutor 
        # it goes from 1 to 10 to 100 etc. based on the height of the branch 
        # can probably do something like str() and converting back to int() as well 
            if node.left: 
                stack.append((node.left, node.value + v * 10)) 
            if node.right: 
                stack.append((node.right, node.value + v * 10)) 
        else: 
            branchesSum += node.value + v * 10 
    return branchesSum