Reverse Only Letters

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

class Solution(object):
    def reverseOnlyLetters(self, S):
        ans = []
        j = len(ans) - 1
        for i, x in enumerate(S):
            if x.isalpha():
                while not S[j].isalpha():
                    j -= 1
                ans.append(S[j])
                j -= 1
            else:
                ans.append(x)

        return "".join(ans)

The function reverseOnlyLetters reverses only the letters in a string, while preserving the positions of non-letter characters. Here’s how it works:

1. The function first creates an empty list ans that will be used to build the result string.

2. The function then sets j to be the last index of the string S.

3. It then iterates over the characters x in S using enumerate to get both the index i and the character x.

4. If x is a letter (checked using x.isalpha()), the function enters a loop where it moves j to the left until it finds a letter in S. It then adds that letter to ans, and moves j one step to the left.

5. If x is not a letter, the function simply adds x to ans.

6. Finally, the function joins the characters in ans together into a string (using "".join(ans)) and returns this string.

So, for example, if the input string S is “ab-cd”, the function will return “dc-ba”. If S is “a-bC-dEf-ghIj”, it will return “j-Ih-gfE-dCba”.

Please note that there’s a typo in the initial j value assignment. It should be j = len(S) - 1, not j = len(ans) - 1, because ans is an empty list at that point, so len(ans) - 1 would always be -1. The corrected code is as follows:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

class Solution(object):
    def reverseOnlyLetters(self, S):
        ans = []
        j = len(S) - 1
        for i, x in enumerate(S):
            if x.isalpha():
                while not S[j].isalpha():
                    j -= 1
                ans.append(S[j])
                j -= 1
            else:
                ans.append(x)

        return "".join(ans)

The function reverseOnlyLetters correctly reverses only the letters in the string S, while preserving the positions of non-letter characters. When tested with the string “a-bC-dEf-ghIj”, it correctly outputs “j-Ih-gfE-dCba”.

Why do we need: while not S[j].isalpha(): j -= 1

The purpose of the line while not S[j].isalpha(): j -= 1 is to find the next letter character from the end of the string S.

When we find a letter at the current position from the start of the string (denoted by i), we want to replace it with a letter from the end of the string (denoted by j). But j might not always point to a letter – it could point to a non-letter character. So, we use a while loop to keep moving j towards the start of the string until we find a letter.

In other words, this loop ensures that j always points to a letter character in the string. This way, when we replace a letter from the start of the string, we are always replacing it with another letter from the end of the string, which is the goal of the reverseOnlyLetters function.

In the context of this problem, j is a read pointer. It reads characters from the end of the string towards the beginning. It’s used to find the next letter character from the end of the string that will replace the current letter character from the beginning of the string.

The read pointer j keeps moving towards the start of the string until it finds a letter. Once it finds a letter, it places that letter in the current position of the output string and then moves one step towards the beginning of the string. This process continues until all letters in the input string have been processed.

So in essence, j acts as a read pointer reading characters in reverse order from the end of the string.

The while loop is there to ensure that when we replace a letter character at the current position in the string, we’re replacing it with another letter character from the end of the string. If j points to a non-letter character, we need to move j towards the start of the string until it points to a letter character.

Let’s take an example to illustrate this. Let’s say S = "a-bC-dEf-ghIj".

As you can see, each time S[i] is a letter, S[j] is also a letter, thanks to the while loop. When S[i] is not a letter, we simply copy it to ans without moving j. This process results in all the letter characters in the string being reversed, while all the non-letter characters remain in their original positions.

A loop invariant is a condition that is initially true and remains true after each iteration of a loop. It’s used to prove the correctness of an algorithm.

In this case, the loop invariant is that at the start of each iteration of the for loop:

1. All positions from 0 to i - 1 in ans have been correctly filled. This means that all the letter characters in these positions have been replaced with letter characters from the end of the string, and all the non-letter characters have been copied as is.

2. j is the index of the next letter character from the end of the string that will be used to replace a letter character at the current position i.

These conditions are true before the first iteration (as ans is empty and j points to the last index of the string), and they remain true after each iteration of the for loop. This guarantees that at the end of the loop, all positions in ans have been correctly filled, giving us the correctly reversed string.

An invariant in computer science and mathematics is a condition that remains unchanged before and after an operation or throughout an algorithm.

In the context of the reverseOnlyLetters function, a suitable invariant would be:

• Non-letter characters remain in their original positions.

This invariant holds true throughout the function. Whether the function is in the process of reversing the letters, or has already done so, non-letter characters always stay in their original positions. This is because the function specifically checks if a character is a letter before deciding whether to reverse it. If a character is not a letter, it is simply appended to the ans list without being reversed.

Therefore, the position of non-letter characters in the string is an invariant in this problem. No matter what else changes, this condition remains the same throughout the execution of the algorithm.

The line of code that maintains the invariant “Non-letter characters remain in their original positions” in the reverseOnlyLetters function is:

1
2

else:
    ans.append(x)

When x (which is S[i]) is not a letter, the function simply appends x to the ans list. This operation keeps the non-letter character x in its original position i. No matter how the letters in the string are reversed, non-letter characters are never moved from their original positions. This line of code is executed each time a non-letter character is encountered, thus maintaining the invariant throughout the execution of the function.

The loop invariant for the function reverseOnlyLetters is: “At the start of each iteration of the for loop, the first i characters in ans are the correct characters that should be at those positions in the final result.”

The following lines of code maintain this loop invariant:

1
2
3
4
5
6
7

if x.isalpha():
    while not S[j].isalpha():
        j -= 1
    ans.append(S[j])
    j -= 1
else:
    ans.append(x)

Here’s how:

• If x (which is S[i]) is a letter, the function uses a while loop to find the next letter character from the end of the string (with j pointing to it). This letter character is then appended to ans, and j is decremented by 1. This ensures that ans now has the correct character at position i.

• If x is not a letter, the function simply appends x to ans. This also ensures that ans now has the correct character at position i.

Therefore, at the end of each iteration of the for loop, the first i + 1 characters in ans are the correct characters that should be at those positions in the final result, which maintains the loop invariant.

Key Takeaway

This problem is a great example for illustrating the importance of maintaining the invariant and loop invariant in the solution.

The problem of reversing only letters in a string serves as a great example to illustrate the significance of maintaining invariants and loop invariants in problem solving.

An invariant, like “Non-letter characters remain in their original positions”, sets a condition that remains unchanged throughout the execution of the algorithm. It provides a constant, reliable condition that we can leverage to solve the problem.

On the other hand, a loop invariant, such as “At the start of each iteration of the for loop, the first i characters in ans are the correct characters that should be at those positions in the final result”, is crucial for understanding the behavior inside a loop and reasoning about its correctness.

By understanding and maintaining these invariants, we are able to devise a solution that systematically swaps the positions of the letter characters while leaving the non-letter characters untouched. This approach ensures that the algorithm works correctly and produces the expected output.

Invariants and loop invariants serve as a powerful tool for designing, understanding, and verifying the correctness of algorithms. They’re essential in many areas of computer science and programming, especially in complex problem solving and algorithmic thinking.

Two Pointers Approach

Here is a Python solution that uses two pointers:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13

class Solution:
    def reverseOnlyLetters(self, s: str) -> str:
        s = list(s)
        left, right = 0, len(s) - 1
        while left < right:
            if not s[left].isalpha():
                left += 1
            elif not s[right].isalpha():
                right -= 1
            else:
                s[left], s[right] = s[right], s[left]
                left, right = left + 1, right - 1
        return "".join(s)

The method reverseOnlyLetters first converts the string s into a list, which makes it easy to swap letters in place.

Then, it initializes two pointers, left and right, at the beginning and end of the list, respectively.

In a loop, the function continues as long as left is less than right:

• If the character at the left index is not an English letter, the function increments left to move the pointer to the next character.
• If the character at the right index is not an English letter, the function decrements right to move the pointer to the previous character.
• If both characters are English letters, the function swaps them and moves both pointers towards the center of the list.

Finally, the function converts the list back into a string and returns it.

This solution uses the isalpha() function to check if a character is an English letter. The isalpha() function returns True if the character is a letter, and False otherwise.

The time complexity of this solution is O(n), where n is the length of the string s, as we are iterating through the string once. The space complexity is also O(n), as we are creating a list to hold the characters of s.

Why do we need two pointers? Why not use one pointer that starts from end of the string and constructs a new string by traversing backwards?

A single pointer traversing from the end of the string and constructing a new string by appending letters in reverse order wouldn’t be sufficient for this problem, because non-letter characters need to remain in their original positions.

If we simply reversed the order of all characters in the string, the positions of these non-letter characters would change. However, the problem requires us to keep the positions of non-letter characters fixed, while only the positions of letters should be reversed.

By using two pointers (one starting from the beginning of the string, and one from the end), we can swap the positions of letter characters while leaving non-letter characters in place. This allows us to reverse the order of the letters without disrupting the positions of the non-letter characters.

For example, consider the string "a-bC-dEf-ghIj". If we just reversed the string, we would get "jIhg-fEd-Cb-a", which doesn’t satisfy the requirement as the positions of the hyphens have been changed. But with two pointers approach, we get "j-Ih-gfE-dCba", where only the positions of the letters are reversed, and the hyphens remain in their original positions.

The problem statement provides a couple of key pieces of information that point us towards the solution:

1. All the English letters (lowercase or uppercase) should be reversed.
2. All the characters that are not English letters remain in the same position.

From these statements, we understand that we need to reverse the positions of the English letters, while maintaining the positions of the non-letter characters. The simplest way to accomplish this is by swapping the positions of the letters.

The concept of ‘reversing’ often involves swapping in algorithms. For instance, to reverse an array, we swap the first and last elements, then the second and penultimate elements, and so on.

Here, we’re doing a similar thing, but we’re only reversing certain elements (the letters), which leads us to the idea of using two pointers - one from the beginning and one from the end, which move towards each other and swap elements as they go.

The constraint that non-letter characters must remain in the same position solidifies this idea, as using two pointers allows us to easily skip these characters and focus on swapping only the letters.