How do you solve Minimum Length of String After Operations on LeetCode?

Minimum Length of String After Operations (LeetCode #3223) 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 characters with a frequency of 3 or more can be removed.

What is the time complexity of Minimum Length of String After Operations?

The optimal solution for Minimum Length of String After Operations runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the string a couple of times to count frequencies and calculate the final length.

What is the brute force solution for Minimum Length of String After Operations?

The brute force approach for Minimum Length of String After Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible character in the string to see if we can perform the operations. We repeatedly remove characters until no more operations can be performed. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Length of String After Operations?

Minimum Length of String After Operations uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Length of String After Operations?

Always think about character frequency when dealing with string manipulation problems. Consider edge cases, such as strings with all unique characters or all identical characters. Explain your thought process clearly to the interviewer, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Minimum Length of String After Operations?

Not counting character frequencies correctly. Assuming characters can be removed without checking their counts.

#3223

Minimum Length of String After Operations

Medium

Hash TableStringCountingHash 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 solution focuses on counting the occurrences of each character. If a character appears at least three times, we can effectively remove it. The final length of the string will be the sum of characters that appear less than three times.

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Algorithm

3 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: For each character, if its frequency is 3 or more, it can be removed completely.
3Step 3: Sum the lengths of characters that appear less than three times to get the minimum length.

solution.py5 lines

1from collections import Counter
2
3def min_length_optimal(s):
4    freq = Counter(s)
5    return sum(count for count in freq.values() if count < 3) + sum(1 for count in freq.values() if count >= 3)

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string a couple of times to count frequencies and calculate the final length.

1Only characters with a frequency of 3 or more can be removed.
2The final length is determined by characters that appear less than 3 times.

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