How do you solve Reverse String Prefix on LeetCode?

Reverse String Prefix (LeetCode #3794) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Reversing a substring can be done in place for efficiency.

What is the time complexity of Reverse String Prefix?

The optimal solution for Reverse String Prefix runs in O(n) time and uses O(n) space. The algorithm runs in linear time as we only traverse the first k characters once.

What is the brute force solution for Reverse String Prefix?

The brute force approach for Reverse String Prefix is Brute Force, with O(n²) time complexity and O(1) space complexity. We can reverse the first k characters by manually swapping each character. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

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

Reverse String Prefix 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 Prefix?

Explain your thought process clearly. Consider edge cases before coding. Optimize your solution after the brute force approach.

What are common mistakes when solving Reverse String Prefix?

Not handling edge cases where k equals the length of the string. Confusing string immutability in certain languages.

#3794

Reverse String Prefix

Easy

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Instead of creating a new string multiple times, we can reverse the first k characters in place and then concatenate the rest, which is more efficient.

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Algorithm

3 steps

1Step 1: Convert the string to a list of characters for mutability.
2Step 2: Use two pointers to swap characters in the first k positions.
3Step 3: Convert the list back to a string and return it.

solution.py4 lines

1def reversePrefix(s, k):
2    s = list(s)
3    s[:k] = s[:k][::-1]
4    return ''.join(s)

ℹ

Complexity note: The algorithm runs in linear time as we only traverse the first k characters once.

1Reversing a substring can be done in place for efficiency.
2Using two pointers is a common technique for swapping elements.

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