How do you solve Apply Operations to Make String Empty on LeetCode?

Apply Operations to Make String Empty (LeetCode #3039) 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: Only the most frequent characters will remain before the last operation.

What is the brute force solution for Apply Operations to Make String Empty?

The brute force approach for Apply Operations to Make String Empty is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly scanning the string to remove the first occurrence of each character from 'a' to 'z'. This is simple to understand but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Apply Operations to Make String Empty?

Apply Operations to Make String Empty uses the following concepts: Array, Hash Table, Sorting, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Apply Operations to Make String Empty?

Always start with a brute-force solution to demonstrate understanding. Think about character frequencies and how they impact the final string. Be prepared to explain your thought process clearly and justify your optimizations.

What are common mistakes when solving Apply Operations to Make String Empty?

Not considering the last occurrences of characters correctly. Failing to optimize the solution, leading to unnecessary complexity.

#3039

Apply Operations to Make String Empty

Medium

ArrayHash TableSortingCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach involves counting the frequency of each character and retaining only the last occurrence of the most frequent characters. This reduces unnecessary operations and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: Identify the maximum frequency of any character.
3Step 3: Construct a new string containing only the last occurrences of characters that have the maximum frequency.

solution.py7 lines

1def apply_operations(s):
2    from collections import Counter
3    freq = Counter(s)
4    max_freq = max(freq.values())
5    last_occurrence = {c: i for i, c in enumerate(s) if freq[c] == max_freq}
6    result = ''.join(c for c in s if c in last_occurrence)
7    return result

ℹ

Complexity note: The time complexity is O(n) because we only pass through the string a few times to count frequencies and build the result. The space complexity is O(n) due to the storage of character frequencies.

1Only the most frequent characters will remain before the last operation.
2The order of characters matters; we need to keep the last occurrence of each character.

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