The Roadmap to Efficient Reversal and Filtering

The method we're discussing here is a universal tool for efficiently reversing any array or string. It's like the Swiss Army knife of reversals. Once you have a grip on this, the next challenge is merely a process of sifting out the non-alphabetical characters, almost like separating wheat from chaff.

Method 1: Isolate, Store, and Reverse

Step 1: Identify Alphabetical Characters

The first step is to identify only the alphabetical characters in the given string. Think of this as a treasure hunt where you only pick up the gold coins and leave the pebbles behind.

Step 2: Store Them Separately

Once you've identified these "gold coins," you store them in a separate array. Imagine placing the gold coins in a separate bucket.

Step 3: Execute the Reversal

After storing these alphabetic characters, the next action is to reverse them. You can use the two-pointer swap method we discussed earlier for this purpose.

Regex to the Rescue

Regular expressions (regex) can be a helpful tool for identifying the alphabetical characters. You can use a pattern like /[a-zA-Z]/ to spot these alphabetic "gold coins" amidst the string of diverse characters.

By following this method, you can isolate what you need, reverse it efficiently, and leave everything else undisturbed.

1r = re.compile("[a-zA-Z]")
2
3for char in s:
4    if r.match(char):
5        alpha_chars.append(char)

The Final Act: Swapping Characters Back

Step 4: Initiate a Second Pass

After reversing our isolated "gold coins" (the alphabetical characters), it's time to go through the string again. This second round, starting from the beginning, has a different objective—swapping.

Step 5: Swap Using Position/Index

The plan is simple: replace the alphabetical characters in the original string with the reversed alphabetical characters from our newly created array, reversedAlpha. Think of it like putting the gold coins back but in a new order.

How It Works

Imagine you have two buckets—one with the original gold coins and the other with the gold coins turned upside down (reversed). As you walk along the path where you initially picked the coins, you place the upside-down coins back in their original spots.

By following these steps, we effectively reverse only the alphabetical characters while keeping the non-alphabetical ones intact. It's as if the coins (alphabets) were flipped, while the pebbles (special characters) remained undisturbed.

1idx_ra = 0
2for i in range(len(s)):
3    if r.match(s[i]):
4        new_s += reversed_alpha[idx_ra]
5        idx_ra += 1
6    else:
7        new_s += s[i]

Navigating the Loop: Advanced Tips

This approach beautifully sidesteps non-alphabetical characters. But let's talk about the "how" for a moment. To ensure accuracy, remember to iterate the reversedAlpha array independently from the loop that scans the original string.

Optional: Simplifying with .shift()

If the thought of juggling another pointer or index leaves you dizzy, here's a simpler alternative: use the .shift() method. This function efficiently grabs the left-most element in an array, essentially doing the pointer's job for you. Imagine it as having a helper who hands you each reversed "gold coin" as you walk along your original path.

Method 2: The Two-Pointer Skip Technique

Dual Pointers with a Twist

The second approach also involves using two pointers, much like our first method. However, this one comes with a twist: it skips over the non-alphabetical characters directly.

How to Skip Efficiently

Just like in a dance, when a couple realizes they are out of sync, they adjust their steps. Similarly, if either of the pointers encounters a non-alphabetical character, it simply "dances" past it, without making a swap.

By adopting this approach, you get to keep the original sequence of non-alphabetical characters intact while reversing only the alphabets. Think of it as flipping some tiles on a mosaic floor without disturbing the overall pattern.

1import re
2
3def swap(arr, a, b):
4    temp = arr[a]
5    arr[a] = arr[b]
6    arr[b] = temp
7    return arr
8
9
10def reverse_only_alpha(s):
11    start = 0
12    end = len(s) - 1
13    arr = list(s)
14
15    r = re.compile("[a-zA-Z]")
16
17    while start <= end:
18        if not r.match(arr[start]):
19            start += 1
20        elif not r.match(arr[end]):
21            end -= 1
22        else:
23            swap(arr, start, end)
24            start += 1
25            end -= 1
26
27    return "".join(arr)
28
29print(reverse_only_alpha("sea!$hells3"))

One Pager Cheat Sheet

• You can reverseOnlyAlphabetical characters in a given string with length <= 100000 and composed of ASCII characters in O(n) time and space complexity.
• The two pointers with swap method is recommended to reverse a regular string.
• We can reverseArray the alphabetical characters of a string using /[a-zA-Z]/ in a for loop to efficiently reverse only the alphabetical characters.
• We can swap out alphabetical characters in the string with elements from our newly created reversedAlpha based on the position/index.
• Using the .map and .filter methods, one can reverse only alphabetical characters in a more succinct manner.
• We split the string into an array and loop over it with map to return non-alphanumeric characters in their current positions.
• We splice and join our array to remove unmatched characters and turn it back into a string, resulting in an O(n) time and space complexity.

This is our final solution.

To visualize the solution and step through the below code, click Visualize the Solution on the right-side menu or the VISUALIZE button in Interactive Mode.

xxxxxxxxxx

    print("Nice job, 3/3 tests passed!")

import re

​

​

def reverse_only_alpha(s):

    alpha_chars = []

    s = list(s)

    new_s = ""

​

    r = re.compile("[a-zA-Z]")

​

    for char in s:

        if r.match(char):

            alpha_chars.append(char)

​

    reversed_alpha = reverse_array(alpha_chars)

​

    idx_ra = 0

    for i in range(len(s)):

        if r.match(s[i]):

            new_s += reversed_alpha[idx_ra]

            idx_ra += 1

        else:

            new_s += s[i]

​

    return new_s

​

​

def reverse_array(arr):

    return arr[::-1]

​

​

import unittest

Got more time? Let's keep going.

If you had any problems with this tutorial, check out the main forum thread here.