How do you solve Reverse String II on LeetCode?

Reverse String II (LeetCode #541) 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: Understanding how to manipulate strings and characters is crucial.

What is the brute force solution for Reverse String II?

The brute force approach for Reverse String II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves reversing the first k characters for every 2k characters in the string. This is done by iterating through the string in chunks and reversing each chunk as needed. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Reverse String II?

Always clarify the requirements and constraints of the problem before jumping into coding. Think aloud during the interview to show your thought process and reasoning. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Reverse String II?

Not considering edge cases where the string length is less than k. Overcomplicating the logic instead of using simple two-pointer techniques.

#541

Reverse String II

Easy

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the string with a two-pointer technique to reverse segments in place, which is more efficient than the brute-force method.

⚙️

Algorithm

5 steps

1Step 1: Convert the string to a list of characters for mutability.
2Step 2: Loop through the string in increments of 2k.
3Step 3: For each segment, reverse the first k characters in place using two pointers.
4Step 4: Continue until the end of the string.
5Step 5: Convert the list back to a string and return it.

solution.py15 lines

1# Full working Python code
2
3def reverseStr(s: str, k: int) -> str:
4    chars = list(s)
5    n = len(chars)
6    for i in range(0, n, 2 * k):
7        left, right = i, min(i + k - 1, n - 1)
8        while left < right:
9            chars[left], chars[right] = chars[right], chars[left]
10            left += 1
11            right -= 1
12    return ''.join(chars)
13
14# Example usage
15print(reverseStr('abcdefg', 2))  # Output: 'bacdfeg'

ℹ

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

1Understanding how to manipulate strings and characters is crucial.
2Recognizing patterns in the problem, such as reversing segments, can lead to efficient solutions.

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