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Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

  1. Two Pointers Approach - Complex approach
  2. Recursive Approach - Complex approach
  3. Iterative Approach with Queue - Complex approach

Approach 1: Two Pointers Approach

Let's start by understanding the problem: we need to repeatedly remove equal characters from both ends of the string until we can't do so anymore, and return the minimum possible length of the resulting string.

Thinking Process: Since we're dealing with operations at both ends of the string, a two-pointer approach seems natural. We can use two pointers, one starting from the beginning of the string (left) and one from the end (right).

If the characters at the left and right pointers are different, we can't perform any operation and the minimum length is the current length of the string. If they are the same, we need to remove all consecutive occurrences of that character from both ends.

After removing these characters, we continue the process with the new ends of the string until either the pointers cross each other or the characters at the two pointers are different.

Intuition: This approach directly simulates the operations described in the problem. By using two pointers, we can efficiently identify and "remove" matching characters from both ends of the string without actually creating new strings in each step.

Algorithm:

  1. Initialize two pointers: left at the beginning of the string and right at the end.
  2. While left is less than right and the characters at left and right are the same:
  3. a. Store the current character.
  4. b. Move left pointer to the right as long as it points to the same character.
  5. c. Move right pointer to the left as long as it points to the same character.
  6. Return the length of the remaining string (right - left + 1).

Time Complexity:

O(n)

Where n is the length of the string. In the worst case, we might need to traverse the entire string once.

Space Complexity:

O(1)

We only use a constant amount of extra space regardless of the input size.

Approach 2: Recursive Approach

We can also solve this problem using a recursive approach, which might be more intuitive for some.

Thinking Process: In each recursive call, we check if the characters at the beginning and end of the string are the same. If they are, we remove all consecutive occurrences of that character from both ends and make a recursive call with the new string.

If the characters are different or the string has a length of 0 or 1, we return the current length of the string as the result.

Intuition: This approach breaks down the problem into smaller subproblems, where each subproblem is to find the minimum length after removing matching characters from both ends of the current string.

Algorithm:

  1. If the string is empty or has only one character, return its length.
  2. If the characters at the beginning and end of the string are different, return the length of the string.
  3. Otherwise, find all consecutive occurrences of the character at the beginning and end of the string.
  4. Remove these characters and make a recursive call with the new string.
  5. Return the result of the recursive call.

Time Complexity:

O(n)

Where n is the length of the string. In the worst case, we might need to traverse the entire string once.

Space Complexity:

O(n)

The recursive approach uses the call stack, which in the worst case might be O(n) deep.

Approach 3: Iterative Approach with Queue

Another approach is to use a queue to simulate the operations.

Thinking Process: We can use a double-ended queue to represent the string. In each iteration, we check if the characters at the front and back of the queue are the same. If they are, we remove all consecutive occurrences of that character from both ends of the queue.

We continue this process until either the queue is empty or the characters at the front and back of the queue are different.

Intuition: This approach is similar to the two-pointer approach but uses a queue data structure to represent the string, which might be more intuitive for some.

Algorithm:

  1. Initialize a double-ended queue with the characters of the string.
  2. While the queue has at least 2 elements and the characters at the front and back of the queue are the same:
  3. a. Store the current character.
  4. b. Remove all consecutive occurrences of that character from the front of the queue.
  5. c. Remove all consecutive occurrences of that character from the back of the queue.
  6. Return the length of the remaining queue.

Time Complexity:

O(n)

Where n is the length of the string. In the worst case, we might need to traverse the entire string once.

Space Complexity:

O(n)

We need to store the characters of the string in the queue, which requires O(n) space.

Pseudocode

solution.pseudo
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function minimumLength(s):
    left = 0
    right = length of s - 1
    
    while left < right and s[left] == s[right]:
        current = s[left]
        
        // Skip all consecutive occurrences of current from the left
        while left < right and s[left] == current:
            left++
        
        // Skip all consecutive occurrences of current from the right
        while left < right and s[right] == current:
            right--
    
    return right - left + 1
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