How do you solve String Compression III on LeetCode?

String Compression III (LeetCode #3163) 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: Count consecutive characters efficiently.

What is the brute force solution for String Compression III?

The brute force approach for String Compression III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through the string and repeatedly finding the longest prefix of the same character, appending its count and character to the result. This method is straightforward but inefficient as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve String Compression III?

String Compression III uses the following concepts: String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for String Compression III?

Explain your thought process clearly. Consider edge cases like single characters. Optimize your solution after the brute-force approach.

What are common mistakes when solving String Compression III?

Not handling the last character group correctly. Forgetting to limit counts to 9.

#3163

String Compression III

Medium

StringTwo 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 single pass through the string to count consecutive characters and build the compressed string. This is efficient because it reduces the number of iterations needed, resulting in a linear time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty result string `comp` and a count variable.
2Step 2: Iterate through the string while counting consecutive characters.
3Step 3: When a different character is encountered or the end of the string is reached, append the count and character to `comp`, and reset the count.

solution.py10 lines

1def compress(word):
2    comp = ''
3    count = 1
4    for i in range(1, len(word) + 1):
5        if i < len(word) and word[i] == word[i - 1]:
6            count += 1
7        else:
8            comp += str(min(count, 9)) + word[i - 1]
9            count = 1
10    return comp

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string. The space complexity is O(n) due to the storage used for the compressed string.

1Count consecutive characters efficiently.
2Use string builders for better performance.

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