Lesson 4

Iterating Through an Array from Middle to Ends

Introduction

Greetings! Welcome to our lesson today, where we'll be taking a deep dive into an intriguing aspect of array manipulation. Have you ever considered the idea of traversing an array not from start to end or from end to start, but from the middle, extending towards both ends simultaneously? Today's lesson is all about this concept. Trust me, it's going to be fascinating! Let's jump right in.

Task Statement

Our task is as follows: Given an array of integers, we aim to return a new array that emerges from the center of the original array and alternates direction towards both ends. In other words, the first element of our new array will be the middle element of the original one. After establishing the starting point, we will alternate between the elements to the left and to the right of the initial center until we have incorporated every element. If the length of the initial array is even, we first take the the element to the left of the center, then the one to the right of the center, then do the alternation as described above.

For example, for numbers = [1, 2, 3, 4, 5], the output should be [3, 2, 4, 1, 5].

This task might initially seem complex. However, don't worry. We're going to break it down and construct our solution step by step. Keep in mind an additional condition: the length of the array—represented as n—can vary from 1 to 100,000, inclusive.

Solution Building: Step 1

To start, we need to establish the middle point of our array. Why the middle, you ask? Well, our task necessitates that we traverse the array from the center to the ends. Python's integer division operator, //, allows us to determine the middle. If the array has an odd length, we add the middle element to the new_order array (as it's not paired with any other element), otherwise we keep it empty for now.

Here's a look at its application in the code:

Python
1def iterateMiddleToEnd(numbers): 2 mid = len(numbers) // 2 # The index of the left middle element 3 if len(numbers) % 2 == 1: 4 new_order = [numbers[mid]] # Adding the middle element to the resulting array 5 else: 6 new_order = [] # No elements in the resulting array for now
Solution Building: Step 2

The resolution of our task requires two pointers: left and right. These pointers will originate from just before and just after the middle element, respectively.

Here's an illustration of how we can implement this in our Python function:

Python
1def iterateMiddleToEnd(numbers): 2 mid = len(numbers) // 2 # The index of the left middle element 3 if len(numbers) % 2 == 1: 4 left = mid - 1 # The left to the middle element 5 right = mid + 1 # The right to the middle element 6 new_order = [numbers[mid]] # Adding the middle element to the resulting array 7 else: 8 left = mid - 1 # Left middle element 9 right = mid # Right middle element 10 new_order = [] # No elements in the resulting array for now
Solution Building: Step 3

Now that our pointers are initialized, it's time to traverse the array and construct our new order. For this, we can effectively utilize a while loop in Python, which continues looping until the left pointer is less than zero and the right pointer is not yet at the end of the list. In each iteration, we add the element at index left to the new_order list, decrease the left pointer by one, add the element at index right, and increase the right pointer by one.

This is what the code looks like:

Python
1def iterateMiddleToEnd(numbers): 2 mid = len(numbers) // 2 # The index of the left middle element 3 if len(numbers) % 2 == 1: 4 left = mid - 1 # The left to the middle element 5 right = mid + 1 # The right to the middle element 6 new_order = [numbers[mid]] # Adding the middle element to the resulting array 7 else: 8 left = mid - 1 # Left middle element 9 right = mid # Right middle element 10 new_order = [] # No elements in the resulting array for now 11 12 while left >= 0 and right < len(numbers): 13 new_order.append(numbers[left]) 14 new_order.append(numbers[right]) 15 left -= 1 16 right += 1 17 18 return new_order

This way, we have created a new ordered array, starting from the middle and alternating between left and right elements all the way to the end of the original array. This approach capably satisfies the requirements of our task!

Lesson Summary

Congratulations on reaching the end of this lesson! You've just uncovered a fascinating method of traversing and manipulating arrays! Be proud of yourself for comprehending this concept and implementing it in the code. As is always the case, practice makes perfect. Therefore, I encourage you to apply this concept to similar problems. Your journey in mastering Python and algorithms has just begun. Take the next step, and code away!

Enjoy this lesson? Now it's time to practice with Cosmo!

Practice is how you turn knowledge into actual skills.