Code Implementation
String End Trimmer Implementation
Below is the implementation of the string end trimmer:
123456789101112131415161718192021222324252627282930/** * Two Pointers Approach * Time Complexity: O(n) - We might need to traverse the entire string once * Space Complexity: O(1) - We only use a constant amount of extra space * * @param {string} s - The input string * @return {number} - The minimum length after performing the operations */function minimumLength(s) { let left = 0; let right = s.length - 1; // Continue until pointers meet or characters at both ends are different while (left < right && s[left] === s[right]) { const current = s[left]; // Skip all consecutive occurrences of current from the left while (left <= right && s[left] === current) { left++; } // Skip all consecutive occurrences of current from the right while (left <= right && s[right] === current) { right--; } } // Return the length of the remaining string return right - left + 1;}Step-by-Step Explanation
Let's break down the implementation:
- 1. Understand the Problem: First, understand that 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.
- 2. Set Up Two Pointers: Initialize two pointers: left at the beginning of the string and right at the end.
- 3. Implement the Two-Pointer Approach: Use a while loop to continue until the pointers meet or the characters at both ends are different.
- 4. Handle Character Removal: When characters at both ends are the same, remove all consecutive occurrences of that character from both ends.
- 5. Calculate the Result: Return the length of the remaining string, which is right - left + 1.
- 6. Handle Edge Cases: Consider edge cases such as when the entire string consists of the same character or when the string has a length of 0 or 1.
- 7. Test the Solution: Verify the solution with the provided examples and additional test cases.
String Problems
Learning Objective
Implement the string end trimmer solution in different programming languages.
String End Trimmer Implementation
Below is the implementation of the string end trimmer in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930/** * Two Pointers Approach * Time Complexity: O(n) - We might need to traverse the entire string once * Space Complexity: O(1) - We only use a constant amount of extra space * * @param {string} s - The input string * @return {number} - The minimum length after performing the operations */function minimumLength(s) { let left = 0; let right = s.length - 1; // Continue until pointers meet or characters at both ends are different while (left < right && s[left] === s[right]) { const current = s[left]; // Skip all consecutive occurrences of current from the left while (left <= right && s[left] === current) { left++; } // Skip all consecutive occurrences of current from the right while (left <= right && s[right] === current) { right--; } } // Return the length of the remaining string return right - left + 1;}Step-by-Step Explanation
11. Understand the Problem
First, understand that 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.
22. Set Up Two Pointers
Initialize two pointers: left at the beginning of the string and right at the end.
33. Implement the Two-Pointer Approach
Use a while loop to continue until the pointers meet or the characters at both ends are different.
44. Handle Character Removal
When characters at both ends are the same, remove all consecutive occurrences of that character from both ends.
55. Calculate the Result
Return the length of the remaining string, which is right - left + 1.
66. Handle Edge Cases
Consider edge cases such as when the entire string consists of the same character or when the string has a length of 0 or 1.
77. Test the Solution
Verify the solution with the provided examples and additional test cases.
Language-Specific Notes
javascript
- JavaScript's strict equality operator (===) is used to compare characters at both ends of the string.
- The solution uses a two-pointer approach to efficiently identify and 'remove' matching characters from both ends of the string.
- The result is calculated as right - left + 1, which represents the length of the remaining string after all operations.
python
- Python provides two implementations: the two-pointer approach and a queue-based approach using collections.deque.
- Python's clean syntax makes the two-pointer approach easy to implement and understand.
- The queue-based approach uses deque's popleft() and pop() methods to efficiently remove elements from both ends of the queue.
java
- Java provides two implementations: the iterative two-pointer approach and a recursive approach.
- Java's charAt() method is used to access characters in the string, as strings in Java are immutable.
- The recursive approach demonstrates how to solve the problem using recursion, but it has a higher space complexity due to the call stack.
cpp
- C++ provides two implementations: the two-pointer approach and a queue-based approach using std::deque.
- C++'s std::deque is used for the queue-based approach, which provides efficient operations at both ends of the queue.
- The solution uses a range-based for loop to initialize the queue with the characters of the string.
Key Takeaways
- The two-pointer approach is an efficient way to solve this problem, with a time complexity of O(n) and a space complexity of O(1).
- The key insight is to identify and 'remove' matching characters from both ends of the string without actually creating new strings in each step.
- When characters at both ends are the same, we need to remove all consecutive occurrences of that character from both ends.
- The operation is greedy - we always want to remove as many characters as possible in each step.
- The problem can also be solved using a recursive approach or a queue-based approach, but the two-pointer approach is the most efficient in terms of space complexity.
- Understanding the problem constraints is important - the string only consists of characters 'a', 'b', and 'c', which simplifies the implementation.
Try It Yourself
Modify the code to implement an alternative approach and test it with the same examples.
Challenge:
Implement a function that solves the string end trimmer problem using a different approach than shown above.
Common Edge Cases
Empty String
If the input string is empty, the minimum length is 0.
Single Character
If the input string has only one character, the minimum length is 1.
All Same Characters
If the entire string consists of the same character, the minimum length is 0 if the length is even, and 1 if the length is odd.
No Matching Ends
If the characters at the beginning and end of the string are different, no operation can be performed, and the minimum length is the length of the string.
Multiple Operations
The problem requires performing the operation multiple times until no more operations can be performed.