What is the time complexity of Reverse Letters Then Special Characters in a String?

The optimal solution for Reverse Letters Then Special Characters in a String runs in O(n) time and uses O(1) space. The complexity is O(n) since we traverse the string once, and we use O(1) space as we modify the string in place.

What is the brute force solution for Reverse Letters Then Special Characters in a String?

The brute force approach for Reverse Letters Then Special Characters in a String is Brute Force, with O(n²) time complexity and O(n) space complexity. This approach involves extracting letters and special characters separately, reversing both lists, and then reconstructing the string. It's straightforward but inefficient due to multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Letters Then Special Characters in a String?

Reverse Letters Then Special Characters in a String uses the following concepts: Two Pointers, String, Simulation. The recommended patterns to study are: Two Pointers, In-place Reversal.

What are the interview tips for Reverse Letters Then Special Characters in a String?

Explain your thought process clearly. Discuss edge cases and how you would handle them. Practice coding without an IDE to simulate interview conditions.

What are common mistakes when solving Reverse Letters Then Special Characters in a String?

Not handling edge cases like empty strings. Confusing letter and special character checks.

#3823

Reverse Letters Then Special Characters in a String

Easy

Two PointersStringSimulationTwo PointersIn-place Reversal

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Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Utilizing two pointers allows us to reverse letters and special characters in a single pass, making it efficient. This method minimizes space usage by avoiding extra lists.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers, one at the start and one at the end of the string.
2Step 2: Move both pointers towards the center, swapping letters and special characters as needed.
3Step 3: Continue until the pointers meet.

solution.py11 lines

1def reverseString(s):
2    s = list(s)
3    left, right = 0, len(s) - 1
4    while left < right:
5        if s[left].isalpha() and s[right].isalpha():
6            s[left], s[right] = s[right], s[left]
7            left += 1
8            right -= 1
9        if not s[left].isalpha(): left += 1
10        if not s[right].isalpha(): right -= 1
11    return ''.join(s)

ℹ

Complexity note: The complexity is O(n) since we traverse the string once, and we use O(1) space as we modify the string in place.

1Two pointers can efficiently handle swaps in-place.
2Reversing in-place avoids additional space usage.

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