algoadvance

You are given a string s consisting only of characters ‘0’ and ‘1’. You can apply the following operation any number of times: Choose any two adjacent characters and remove them if they are different (i.e. “01” or “10”). This operation reduces the string’s length by 2. Your task is to return the minimum possible length of the string after performing the operations as many times as possible.

Clarifying Questions

  1. What if the string is empty initially?
    • If the string is empty, the minimum length is 0 by default.
  2. Are there any constraints on the length of the string?
    • Typical constraints should be provided in the problem statement but for now, assume it can be reasonably large, up to around (10^5) characters.
  3. What should be the format of the output?
    • The output should be a single integer indicating the minimum possible length of the string.

Strategy

The strategy to solve this problem is to simulate the removal operation using a stack-based approach. The basic idea is as follows:

  1. Traverse through each character in the string.
  2. Use a stack to keep track of characters.
  3. For each character, if the stack is not empty and the top of the stack is different from the current character, we can remove (pop) the top of the stack because they form a “01” or “10” pair.
  4. If not, we push the current character onto the stack.
  5. By the end of the traversal, the stack will contain characters that could not be removed. The length of this stack will be our answer.

Code

def minLengthAfterRemovals(s: str) -> int:
    stack = []
    
    for char in s:
        if stack and stack[-1] != char:
            stack.pop()  # This removes a "01" or "10" pair.
        else:
            stack.append(char)  # Push current character onto stack.
    
    return len(stack)

# Example usage
s = "010101"
print(minLengthAfterRemovals(s))  # Output: 0

Time Complexity

The time complexity of this approach is (O(n)), where (n) is the length of the string. This is because we are traversing the string once and each push or pop operation on the stack takes (O(1)) time.

The space complexity is also (O(n)) in the worst case, where no characters are removable, and all characters are stored in the stack.

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