How do you solve Delete Characters to Make Fancy String on LeetCode?

Delete Characters to Make Fancy String (LeetCode #1957) 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: To maintain a fancy string, we only need to ensure that no character appears three times consecutively.

What is the brute force solution for Delete Characters to Make Fancy String?

The brute force approach for Delete Characters to Make Fancy String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of the string and removing characters until no three consecutive characters are the same. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Delete Characters to Make Fancy String?

Delete Characters to Make Fancy String uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Delete Characters to Make Fancy String?

Explain your thought process clearly as you work through the problem. Start with a brute-force solution to show your understanding before optimizing. Be mindful of edge cases, such as strings that are already fancy or consist of the same character.

What are common mistakes when solving Delete Characters to Make Fancy String?

Not resetting the count correctly when encountering a different character. Overcomplicating the logic by trying to manipulate substrings instead of building the result directly.

#1957

Delete Characters to Make Fancy String

Easy

StringTwo PointersSliding Window

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 approach uses a single pass through the string while keeping track of consecutive characters. This allows us to build the result string efficiently without unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result string and a count variable.
2Step 2: Iterate through each character of the input string.
3Step 3: For each character, check if it is the same as the last character added to the result. If it is, increase the count; otherwise, reset the count.
4Step 4: If the count is less than 3, append the character to the result.

solution.py11 lines

1def makeFancyString(s):
2    result = ''
3    count = 1
4    for i in range(len(s)):
5        if i > 0 and s[i] == s[i - 1]:
6            count += 1
7        else:
8            count = 1
9        if count < 3:
10            result += s[i]
11    return result

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(n) due to the storage of the result string.

1To maintain a fancy string, we only need to ensure that no character appears three times consecutively.
2Using a single pass through the string is more efficient than checking all substrings.

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