Reverse String Leetcode Problem 344 [Python Solution]

Welcome to another coding challenge!

In this post, we'll tackle the Reverse String problem, which is categorized under the Two Pointers technique.

This problem is a great fit for beginners, as it offers multiple ways to solve it, and we'll explore these approaches in detail.

We'll not only provide a Python solution but also discuss different strategies, including the efficient use of space and time complexity.

Let's get started by understanding the problem and its constraints.

Problem Overview

The problem is as follows:

Question: Write a function that reverses a string.

The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

Constraints:

• 1 <= s.length <= 105
• s[i] is a printable ASCII character.

Now, let's dive into a more detailed exploration of the problem and its associated constraints.

Understanding Constraints

The problem's constraints provide valuable insights into how we should approach solving it.

Let's break down these constraints:

1. 1 <= s.length <= 105: This tells us that the input string can be quite long, but it will not exceed 105 characters.

2. s[i] is a printable ASCII character: This constraint specifies that the characters in the input string s are limited to printable ASCII characters, which include letters, digits, and symbols.

Reverse String LeetCode Problem Solution

Bruteforce Approach

One way to solve the problem is through a simple brute-force approach.

The idea is to swap the characters from the two ends of the array until the entire string is reversed.

Here's the Python code for this approach:

def reverseString(self, s: List[str]) -> None:
    """
    Do not return anything, modify s in-place instead.
    """
    l = 0
    r = len(s) - 1
    while l < r:
        s[l], s[r] = s[r], s[l]
        l += 1
        r -= 1

In this code, we maintain two pointers, l and r, representing the left and right ends of the string, respectively.

We swap the characters at these positions and continue moving towards the center until the entire string is reversed.

This approach has a time complexity of O(n), where n is the length of the input string s.

The space complexity is O(1) because we modify the input array in-place without using any additional data structures.

Efficient Approach with Two Pointers

The brute-force approach is efficient, but let's explore a more space-efficient method using the two-pointer technique.

This approach is not only memory-efficient but also easy to understand.

In this approach, we use two pointers: one at the beginning (left) and one at the end (right) of the string.

We swap the characters at these pointers and then increment the left pointer and decrement the right pointer.

We repeat this process until the two pointers meet in the middle.

This approach also has a time complexity of O(n) and, more importantly, a space complexity of O(1), as we perform all the operations in-place without using any extra memory.

Now, let's move on to the code for this efficient approach.

def reverseString(self, s: List[str]) -> None:
    """
    Do not return anything, modify s in-place instead.
    """
    l = 0
    r = len(s) - 1
    while l < r:
        s[l], s[r] = s[r], s[l]
        l += 1
        r -= 1

The code is nearly identical to the brute-force approach, but it's worth emphasizing the space efficiency, which makes it a preferred solution.

Time and Space Complexity

Let's summarize the time and space complexities of the two approaches:

• Bruteforce Approach:

  • Time Complexity: O(n)
  • Space Complexity: O(1)

• Efficient Approach with Two Pointers:

  • Time Complexity: O(n)
  • Space Complexity: O(1)

Both approaches have the same time complexity, but the efficient approach shines in terms of space efficiency as it uses only constant extra memory.

Reasoning Behind Our Approach

The reason behind using the two-pointer technique is its efficiency.

By maintaining two pointers that traverse the string from both ends, we can achieve the desired outcome without using any additional data structures.

This results in a space-efficient solution, which is essential when solving coding challenges with strict memory constraints.

In the efficient approach, we start with two pointers at opposite ends and gradually swap characters while moving toward the center of the string.

This method is not only simple to implement but also optimally utilizes the available memory.

It allows us to reverse the string with minimal space complexity.

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Conclusion

In this blog post, we tackled the Reverse String problem from LeetCode.

We discussed various approaches, with a focus on space efficiency.

The efficient approach using two pointers is not only easy to understand but also optimized for memory usage, making it a preferred solution.

Remember, as a beginner, it's crucial to explore different problem-solving techniques and understand the constraints provided in the problem statement.

This practice will help you improve your coding skills and prepare for coding interviews.

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We encourage you to ask questions, make suggestions, and share your thoughts in the comments.

Your feedback is valuable!

Click here to view the problem on LeetCode

Happy coding!