What is the time complexity of Minimum Length of String After Deleting Similar Ends?

The optimal solution for Minimum Length of String After Deleting Similar Ends runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the string once with the two pointers, making it linear in relation to the length of the string.

What is the brute force solution for Minimum Length of String After Deleting Similar Ends?

The brute force approach for Minimum Length of String After Deleting Similar Ends is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible prefixes and suffixes to remove them iteratively until no more can be removed. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Length of String After Deleting Similar Ends?

Minimum Length of String After Deleting Similar Ends uses the following concepts: Two Pointers, String. The recommended patterns to study are: Two Pointers, String Manipulation.

What are the interview tips for Minimum Length of String After Deleting Similar Ends?

Clearly explain your thought process and approach. Consider edge cases and test your solution with them. Practice similar problems to become familiar with patterns.

What are common mistakes when solving Minimum Length of String After Deleting Similar Ends?

Not handling edge cases where the string length is 1. Failing to check for non-matching characters before removing.

#1750

Minimum Length of String After Deleting Similar Ends

Medium

Two PointersStringTwo PointersString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a two-pointer technique to efficiently find and remove matching characters from both ends of the string. This approach minimizes unnecessary iterations and directly targets the characters to be removed.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, one at the start (left) and one at the end (right) of the string.
2Step 2: Move the left pointer to the right as long as it points to the same character as the first character.
3Step 3: Move the right pointer to the left as long as it points to the same character as the last character.
4Step 4: If the characters at the left and right pointers are the same, remove them and repeat the process.
5Step 5: Return the length of the remaining string.

solution.py8 lines

1def minimumLength(s):
2    left, right = 0, len(s) - 1
3    while left < right and s[left] == s[right]:
4        while left < right and s[left] == s[0]:
5            left += 1
6        while left < right and s[right] == s[-1]:
7            right -= 1
8    return right - left + 1

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with the two pointers, making it linear in relation to the length of the string.

1Matching characters at both ends can be removed simultaneously.
2The two-pointer technique is efficient for problems involving string manipulation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.