• Nov 19, 2022 •CodeCatch
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# Python program for Plotting Fibonacci # spiral fractal using Turtle import turtle import math def fiboPlot(n): a = 0 b = 1 square_a = a square_b = b # Setting the colour of the plotting pen to blue x.pencolor("blue") # Drawing the first square x.forward(b * factor) x.left(90) x.forward(b * factor) x.left(90) x.forward(b * factor) x.left(90) x.forward(b * factor) # Proceeding in the Fibonacci Series temp = square_b square_b = square_b + square_a square_a = temp # Drawing the rest of the squares for i in range(1, n): x.backward(square_a * factor) x.right(90) x.forward(square_b * factor) x.left(90) x.forward(square_b * factor) x.left(90) x.forward(square_b * factor) # Proceeding in the Fibonacci Series temp = square_b square_b = square_b + square_a square_a = temp # Bringing the pen to starting point of the spiral plot x.penup() x.setposition(factor, 0) x.seth(0) x.pendown() # Setting the colour of the plotting pen to red x.pencolor("red") # Fibonacci Spiral Plot x.left(90) for i in range(n): print(b) fdwd = math.pi * b * factor / 2 fdwd /= 90 for j in range(90): x.forward(fdwd) x.left(1) temp = a a = b b = temp + b # Here 'factor' signifies the multiplicative # factor which expands or shrinks the scale # of the plot by a certain factor. factor = 1 # Taking Input for the number of # Iterations our Algorithm will run n = int(input('Enter the number of iterations (must be > 1): ')) # Plotting the Fibonacci Spiral Fractal # and printing the corresponding Fibonacci Number if n > 0: print("Fibonacci series for", n, "elements :") x = turtle.Turtle() x.speed(100) fiboPlot(n) turtle.done() else: print("Number of iterations must be > 0")
• May 31, 2023 •CodeCatch
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my_list = [1, 2, 3, 4, 5] removed_element = my_list.pop(2) # Remove and return element at index 2 print(removed_element) # 3 print(my_list) # [1, 2, 4, 5] last_element = my_list.pop() # Remove and return the last element print(last_element) # 5 print(my_list) # [1, 2, 4]
• May 5, 2026 •CodeCatch
{"output":{"name":"format_final_json_response","arguments":{"output":{}}}}
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# Python program to reverse a linked list # Time Complexity : O(n) # Space Complexity : O(n) as 'next' #variable is getting created in each loop. # Node class class Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = None class LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to reverse the linked list def reverse(self): prev = None current = self.head while(current is not None): next = current.next current.next = prev prev = current current = next self.head = prev # Function to insert a new node at the beginning def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node # Utility function to print the linked LinkedList def printList(self): temp = self.head while(temp): print temp.data, temp = temp.next # Driver program to test above functions llist = LinkedList() llist.push(20) llist.push(4) llist.push(15) llist.push(85) print "Given Linked List" llist.printList() llist.reverse() print "\nReversed Linked List" llist.printList()
• Oct 15, 2022 •CodeCatch
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my_list = ["blue", "red", "green"] #1- Using sort or srted directly or with specifc keys my_list.sort() #sorts alphabetically or in an ascending order for numeric data my_list = sorted(my_list, key=len) #sorts the list based on the length of the strings from shortest to longest. # You can use reverse=True to flip the order #2- Using locale and functools import locale from functools import cmp_to_key my_list = sorted(my_list, key=cmp_to_key(locale.strcoll))
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list_1 = [1,2,3,4,5,6,7,8,9] cubed = map(lambda x: pow(x,3), list_1) print(list(cubed)) #Results #[1, 8, 27, 64, 125, 216, 343, 512, 729]