import re

def find_pattern(text, pattern):
    match = re.search(pattern, text)
    if match:
        return match.group()
    else:
        return 'No match found'

text = 'Hello, world! This is a test.'
pattern = r'world!.*test.'

result = find_pattern(text, pattern)
print(result)

#Python 3: Fibonacci series up to n
def fib(n):
 a, b = 0, 1
 while a < n:
     print(a, end=' ')
     a, b = b, a+b
 print()
fib(1000)

from time import sleep

def delay(fn, ms, *args):
  sleep(ms / 1000)
  return fn(*args)

delay(lambda x: print(x), 1000, 'later') # prints 'later' after one second

{"output":{"name":"format_final_json_response","arguments":{"output":{}}}}

class Solution(object):
    def floodFill(self, image, sr, sc, newColor):
        R, C = len(image), len(image[0])
        color = image[sr][sc]
        if color == newColor: return image
        def dfs(r, c):
            if image[r][c] == color:
                image[r][c] = newColor
                if r >= 1: dfs(r-1, c)
                if r+1 < R: dfs(r+1, c)
                if c >= 1: dfs(r, c-1)
                if c+1 < C: dfs(r, c+1)

        dfs(sr, sc)
        return image

import anytree as at
import random as rm

# Generate a tree with node_count many nodes. Each has a number key that shows when it was made and a randomly selected color, red or white.
def random_tree(node_count):
    
    # Generates the list of nodes
    nodes = []
    for i in range(node_count):
        test = rm.randint(1,2)
        if test == 1:
            nodes.append(at.Node(str(i),color="white"))
        else:
            nodes.append(at.Node(str(i),color="red"))
    
    #Creates the various main branches
    for i in range(node_count):
        for j in range(i, len(nodes)):
            test = rm.randint(1,len(nodes))
            if test == 1 and nodes[j].parent == None and (not nodes[i] == nodes[j]):
                nodes[j].parent = nodes[i]
    
    #Collects all the main branches into a single tree with the first node being the root
    for i in range(1, node_count):
        if nodes[i].parent == None and (not nodes[i] == nodes[0]):
            nodes[i].parent = nodes[0]

    return nodes[0]