#3223

Medium

LeetCode Problem Workspace

Minimum Length of String After Operations

Find the minimum length of a string after performing operations based on character frequency.

Problem Statement

You are given a string s consisting of lowercase English letters. You can perform the following operation as many times as you want: remove any pair of adjacent characters that are the same. Return the minimum possible length of the string after performing all possible operations.

In each operation, you can only remove adjacent matching pairs, and the frequency of each character determines if a pair can be removed. If no more pairs can be removed, return the length of the resulting string.

Examples

Example 1

Input: s = "abaacbcbb"

Output: 5

We do the following operations:

Example 2

Input: s = "aa"

Output: 2

We cannot perform any operations, so we return the length of the original string.

Constraints

1 <= s.length <= 2 * 105
s consists only of lowercase English letters.

Solution Approach

Track character frequencies

Use a hash table to count the frequency of each character in the string. This will help in determining which characters can potentially form pairs for removal.

Process string for adjacent pairs

Iterate through the string and remove adjacent pairs of characters. Update the string after each removal to continually minimize the length.

Final string length determination

After all removals, the remaining string's length will be the answer. This can be computed by comparing the frequency of characters after all possible pair removals.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity is O(n) since we iterate through the string a constant number of times, and space complexity is O(1) as we only use a fixed amount of space for counting characters.

What Interviewers Usually Probe

Can the candidate quickly implement a solution using a hash table to count frequencies?
Does the candidate consider all edge cases like an already minimal string or no removable pairs?
Is the candidate aware of the importance of focusing only on character frequencies for an optimized solution?

Common Pitfalls or Variants

Common pitfalls

Ignoring edge cases where no adjacent pairs exist and not optimizing for a direct frequency-based solution.
Incorrectly handling cases with an odd number of identical characters that can’t fully form pairs.
Overcomplicating the solution by attempting to manually remove pairs instead of focusing on character frequencies.

Follow-up variants

What happens if the string is composed entirely of unique characters?
Can this problem be solved using a stack approach instead of hash tables?
How does the time complexity change when considering strings of varying lengths or different characters?

FAQ

How do hash tables help in solving the Minimum Length of String After Operations problem?

Hash tables store the frequency of each character, which is essential for determining if adjacent pairs can be removed. This allows for an efficient solution.

What are the time and space complexities for this problem?

The time complexity is O(n), and the space complexity is O(1), as only a fixed amount of space is needed for frequency counting.

What’s the key to efficiently solving this problem?

The key is focusing on character frequencies and using a hash table to track them, minimizing the string through pair removal based on these frequencies.

Can this problem be solved using a stack approach?

Yes, a stack could potentially be used to remove adjacent pairs, but a hash table-based approach is more efficient in terms of frequency counting.

How do I handle edge cases in the Minimum Length of String After Operations?

Edge cases include strings that already have no adjacent pairs or strings where characters cannot fully form pairs. Ensure you account for these scenarios during implementation.

terminal

Solution

Solution 1: Counting

We can count the occurrences of each character in the string, and then iterate through the count array. If a character appears an odd number of times, then $1$ of that character remains in the end; if a character appears an even number of times, then $2$ of that character remain. We can sum the remaining counts of all characters to get the final length of the string.

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class Solution:
    def minimumLength(self, s: str) -> int:
        cnt = Counter(s)
        return sum(1 if x & 1 else 2 for x in cnt.values())

Continue Practicing

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