Is Minimum String Length After Balanced Removals easy or hard?

The problem is typically classified as Medium because the optimal solution requires recognizing the frequency imbalance insight. A straightforward simulation is easy to implement, but deriving the counting formula requires deeper reasoning about how removals interact.

Minimum String Length After Balanced Removals Python/Java solution

Most implementations compute character frequencies using a dictionary or array, then apply the formula based on the maximum frequency and total length. The logic is identical across Python, Java, C++, Go, and TypeScript and runs in linear time.

How to solve Minimum String Length After Balanced Removals in O(n)?

Iterate through the string and count the frequency of each character. Let n be the string length and maxFreq be the highest frequency. If maxFreq > n - maxFreq, the remaining length is maxFreq - (n - maxFreq). Otherwise, the result is n % 2 because balanced removals cancel nearly all characters.

What is the best approach for Minimum String Length After Balanced Removals?

The most efficient solution uses frequency counting. Count how many times each character appears and track the maximum frequency. If the most frequent character exceeds the count of all other characters combined, some characters cannot be paired and remain. Otherwise, nearly all characters cancel out, leaving either 0 or 1 depending on whether the total length is even or odd. This runs in O(n) time and O(1) space.

Is Minimum String Length After Balanced Removals asked at Google/Amazon/Meta?

Problems based on greedy counting and string cancellation patterns frequently appear in interviews at companies like Google, Amazon, and Meta. Variants involving removing pairs, balancing characters, or minimizing remaining length are common interview exercises.

What data structure is used in Minimum String Length After Balanced Removals?

The optimal approach uses a frequency counter, typically implemented with a fixed-size array or hash map. A stack can also be used in a simulation approach to model pair removals during traversal, though it requires more memory.

What is the time complexity of Minimum String Length After Balanced Removals?

The optimal counting solution runs in O(n) time because it scans the string once to compute character frequencies. Space complexity is O(1) since the alphabet size is fixed and only a small frequency array or map is required.

Description Hints 3 Solution Video 2:05

3746. Minimum String Length After Balanced Removals

Medium77.6% acceptance

You are given a string s consisting only of the characters 'a' and 'b'.

You are allowed to repeatedly remove any substring where the number of 'a' characters is equal to the number of 'b' characters. After each removal, the remaining parts of the string are concatenated together without gaps.

Return an integer denoting the minimum possible length of the string after performing any number of such operations.

Example 1:

Input: s = "aabbab"

Output: 0

Explanation:

The substring "aabbab" has three 'a' and three 'b'. Since their counts are equal, we can remove the entire string directly. The minimum length is 0.

Example 2:

Input: s = "aaaa"

Output: 4

Explanation:

Every substring of "aaaa" contains only 'a' characters. No substring can be removed as a result, so the minimum length remains 4.

Example 3:

Input: s = "aaabb"

Output: 1

Explanation:

First, remove the substring "ab", leaving "aab". Next, remove the new substring "ab", leaving "a". No further removals are possible, so the minimum length is 1.

Constraints:

1 <= s.length <= 10⁵
s[i] is either 'a' or 'b'.

Topics

Hints open one at a time

Hint 1

Remove the longest possible balanced substring initially

Hint 2

Let the count of a's be count_a and the count of b's be count_b. Can we derive the final length from these?

Hint 3

The answer is abs(count_a - count_b)

2:05

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1class Solution:
2    def minLengthAfterRemovals(self, s: str) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"aabbab"

Input

"aaaa"

Input

"aaabb"

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