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![Give a step-by-step analysis of the worst-case time complexity of the bubble sort algorithm, with Python code examples, but write every sentence in the speaking style of a fast-talkin' wise guy from a 1940's gangster movie. Alright, pal, listen up. The bubble sort algorithm, she's a real simple one. But the worst- case time complexity, that's a whole different story. You see, in the worst case, every element in the list gotta be swapped, so the time complexity is gonna be proportional to the square of the length of the list. Get it? Here's some Python code to help you out, see if you can follow along. # Start with a list of numbers that ain't sorted numbers = [5, 1, 4, 2, 8] #Keep track of whether any swaps were made on the previous iteration # If no swaps were made, the list is already sorted and we can stop swapped True Copy code while swapped: # Set swapped to False so we can check if any swaps are made later swapped = False # For each pair of adjacent elements, for i in range(len(numbers) - 1): # If the first element is greater than the second, if numbers[i] > numbers [i+1]: #Swap the elements numbers[i], numbers[i+1] = numbers [i+1], numbers[i] # Set swapped to True so we know to keep looping swapped = True # At this point, the list is sorted print(numbers) See, in the worst case, the while loop is gonna keep loopin' until there ain't no more swaps to be made. And the inner for loop, it's gonna go through every element in the list, so the time complexity is gonna be proportional to the square of the length of the list. Get it?](https://i.kym-cdn.com/photos/images/masonry/002/488/215/35b.png)
![Give a step-by-step analysis of the worst-case time complexity of the bubble sort algorithm, with Python code examples, but write every sentence in the speaking style of a fast-talkin' wise guy from a 1940's gangster movie. Alright, pal, listen up. The bubble sort algorithm, she's a real simple one. But the worst- case time complexity, that's a whole different story. You see, in the worst case, every element in the list gotta be swapped, so the time complexity is gonna be proportional to the square of the length of the list. Get it? Here's some Python code to help you out, see if you can follow along. # Start with a list of numbers that ain't sorted numbers = [5, 1, 4, 2, 8] #Keep track of whether any swaps were made on the previous iteration # If no swaps were made, the list is already sorted and we can stop swapped True Copy code while swapped: # Set swapped to False so we can check if any swaps are made later swapped = False # For each pair of adjacent elements, for i in range(len(numbers) - 1): # If the first element is greater than the second, if numbers[i] > numbers [i+1]: #Swap the elements numbers[i], numbers[i+1] = numbers [i+1], numbers[i] # Set swapped to True so we know to keep looping swapped = True # At this point, the list is sorted print(numbers) See, in the worst case, the while loop is gonna keep loopin' until there ain't no more swaps to be made. And the inner for loop, it's gonna go through every element in the list, so the time complexity is gonna be proportional to the square of the length of the list. Get it?](https://i.kym-cdn.com/photos/images/original/002/488/215/35b.png)
