How do you solve Reverse String on LeetCode?

Reverse String (LeetCode #344) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using in-place operations saves memory and is often required in interviews.

What is the brute force solution for Reverse String?

The brute force approach for Reverse String is Brute Force, with O(n²) time complexity and O(1) space complexity. The simplest way to reverse a string is to create a new string and fill it with characters from the original string in reverse order. This is straightforward but inefficient in terms of space. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse String?

Reverse String uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Reverse String?

Always clarify if in-place modification is required before starting. Think about edge cases, such as strings with only one character or an empty string. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Reverse String?

Not modifying the input array in-place as required. Confusing the indices when swapping characters.

#344

Reverse String

Easy

Two PointersStringTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Using the two-pointer technique allows us to reverse the string in-place without needing extra space for a new string. This is efficient and meets the problem's constraints.

⚙️

Algorithm

4 steps

1Step 1: Initialize two pointers, one at the start (left) and one at the end (right) of the array.
2Step 2: Swap the characters at the left and right pointers.
3Step 3: Move the left pointer one step to the right and the right pointer one step to the left.
4Step 4: Repeat until the left pointer is greater than or equal to the right pointer.

solution.py6 lines

1def reverseString(s):
2    left, right = 0, len(s) - 1
3    while left < right:
4        s[left], s[right] = s[right], s[left]
5        left += 1
6        right -= 1

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once. The space complexity is O(1) since we are reversing the string in-place without using extra space.

1Using in-place operations saves memory and is often required in interviews.
2The two-pointer technique is a common pattern for problems involving arrays or strings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.