How do you solve Reverse Words in a String III on LeetCode?

Reverse Words in a String III (LeetCode #557) 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 words can be efficiently done using in-place techniques.

What is the brute force solution for Reverse Words in a String III?

The brute force approach for Reverse Words in a String III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves reversing each word individually by iterating through the string multiple times. This is straightforward but inefficient because it requires extra passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Words in a String III?

Reverse Words in a String III uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Reverse Words in a String III?

Always clarify the problem statement and constraints before diving into coding. Think about the time and space complexity of your solution and be ready to explain it. Practice dry-running your code with sample inputs to catch errors early.

What are common mistakes when solving Reverse Words in a String III?

Not considering edge cases like single-word strings. Overcomplicating the problem by using unnecessary data structures.

#557

Reverse Words in a String III

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)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a two-pointer technique to reverse the words in place. This is efficient because it only requires a single pass through the string and uses minimal extra space.

⚙️

Algorithm

3 steps

1Step 1: Convert the string to a character array for in-place modification.
2Step 2: Use two pointers to identify the start and end of each word, and reverse the characters in that range.
3Step 3: Continue until all words are reversed, then return the modified character array as a string.

solution.py12 lines

1# Full working Python code
2s = "Let's take LeetCode contest"
3char_list = list(s)
4start = 0
5for end in range(len(char_list)):
6    if char_list[end] == ' ' or end == len(char_list) - 1:
7        if end == len(char_list) - 1:
8            end += 1
9        char_list[start:end] = char_list[start:end][::-1]
10        start = end + 1
11result = ''.join(char_list)
12print(result)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string a single time to reverse the words. The space complexity is O(n) due to the character array used for in-place modification.

1Reversing words can be efficiently done using in-place techniques.
2Understanding string manipulation and character arrays is crucial for optimal solutions.

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