How do you solve Reverse Only Letters on LeetCode?

Reverse Only Letters (LeetCode #917) 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: Use two pointers to efficiently reverse elements in place.

What is the brute force solution for Reverse Only Letters?

The brute force approach for Reverse Only Letters is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves reversing only the letters in the string while leaving non-letter characters in their original positions. This can be done by creating a new string and filling it with reversed letters at the appropriate positions. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Only Letters?

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

What are the interview tips for Reverse Only Letters?

Explain your thought process clearly while coding. Consider edge cases, such as strings with no letters or all letters. Practice similar problems to get comfortable with two-pointer techniques.

What are common mistakes when solving Reverse Only Letters?

Not handling non-letter characters correctly. Confusing the order of letters during swapping.

#917

Reverse Only Letters

Easy

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a two-pointer technique to reverse the letters in place. This method is efficient because it only requires a single pass through the string, making it faster than the brute-force method.

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Algorithm

4 steps

1Step 1: Initialize two pointers, one at the start and one at the end of the string.
2Step 2: Move the left pointer right until it points to a letter, and the right pointer left until it points to a letter.
3Step 3: Swap the letters at the two pointers, then move both pointers inward.
4Step 4: Repeat the process until the left pointer is greater than or equal to the right pointer.

solution.py13 lines

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

ℹ

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

1Use two pointers to efficiently reverse elements in place.
2Non-letter characters should be skipped while processing.

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