README.md

Array Reverse

401 Class 01

Whiteboard Process

• v.2 Reference: class 02 code review with JB

Approach & Efficiency

Time

• Go through the number of half the list items to get to midpoint time depends on how big the list is single step operations so operations do not depend on how big the array is (O of 1 operations) swapping incrementing/modifying the right and left indices
• .5 n since go through half the list
• Big O notation, drop the .5
• n since time is related to n (the length of n/the items)
• Linear time complexity

Space

• Not creating a new array
• a little bit of memory allocated but constant since only 1 left and 1 right index no matter how long the array is
• O of 0
• The algorithm uses a fixed amount of space regardless of the input size

Time Complexity: O(N) Space Complexity: O(1)

Solution

def reverse_list(items):
  left_index = 0
  right_index = len(items) - 1
  while left_index < right_index:
    items[left_index], items[right_index] = items[right_index], items[left_index]
    left_index += 1
    right_index -= 1

  return items