How do you solve Reverse Prefix of Word on LeetCode?

Reverse Prefix of Word (LeetCode #2000) 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: Finding the index of the character is crucial for both approaches.

What is the brute force solution for Reverse Prefix of Word?

The brute force approach for Reverse Prefix of Word is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves finding the first occurrence of the character 'ch' and then reversing the substring from the start of 'word' to that index. This is simple 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 Prefix of Word?

Reverse Prefix of Word uses the following concepts: Two Pointers, String, Stack. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Reverse Prefix of Word?

Always clarify the requirements before starting to code. Think about edge cases, such as when 'ch' is not present. Discuss your thought process with the interviewer as you code.

What are common mistakes when solving Reverse Prefix of Word?

Not checking if the character exists before attempting to reverse. Confusing string indexing, especially with inclusive/exclusive boundaries.

#2000

Reverse Prefix of Word

Easy

Two PointersStringStackTwo 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)

The optimal approach uses a single pass to find the index of 'ch' and then constructs the result in one go, making it more efficient. This reduces the number of operations needed to reverse the substring.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to store the index of the first occurrence of 'ch'.
2Step 2: Loop through 'word' to find the index of 'ch'.
3Step 3: If 'ch' is not found, return 'word' as is.
4Step 4: Create a new string by reversing the substring from index 0 to the found index and appending the rest of 'word'.
5Step 5: Return the resulting string.

solution.py5 lines

1def reversePrefix(word, ch):
2    index = word.find(ch)
3    if index == -1:
4        return word
5    return ''.join(reversed(word[:index + 1])) + word[index + 1:]

ℹ

Complexity note: The time complexity is O(n) because we traverse the string only once to find 'ch' and construct the result. The space complexity is O(n) due to the new string being created.

1Finding the index of the character is crucial for both approaches.
2Reversing a substring can be done efficiently using built-in functions.

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