#1750

Medium

auto_awesomeTwo-pointer scanning with invariant tracking

LeetCode Problem Workspace

Minimum Length of String After Deleting Similar Ends

Minimize string length by deleting similar characters from both ends repeatedly using a two-pointer technique.

Problem Statement

You are given a string s consisting only of the characters 'a', 'b', and 'c'. The task is to repeatedly remove characters from both ends of the string. At each step, you remove matching characters from the prefix and suffix of the string, if they are identical. The goal is to return the minimum length of the string after applying this operation any number of times.

For example, in the case of s = "cabaabac", an optimal sequence of operations removes characters from both ends in stages until the string is empty, resulting in a final length of 0. If no characters can be removed, the string's length remains unchanged.

Examples

Example 1

Input: s = "ca"

Output: 2

You can't remove any characters, so the string stays as is.

Example 2

Input: s = "cabaabac"

Output: 0

An optimal sequence of operations is:

Take prefix = "c" and suffix = "c" and remove them, s = "abaaba".
Take prefix = "a" and suffix = "a" and remove them, s = "baab".
Take prefix = "b" and suffix = "b" and remove them, s = "aa".
Take prefix = "a" and suffix = "a" and remove them, s = "".

Example 3

Input: s = "aabccabba"

Output: 3

An optimal sequence of operations is:

Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb".
Take prefix = "b" and suffix = "bb" and remove them, s = "cca".

Constraints

1 <= s.length <= 105
s only consists of characters 'a', 'b', and 'c'.

Solution Approach

Two-pointer approach

Use two pointers: one starting from the leftmost character and the other from the rightmost. Compare the characters at these positions and, if they are the same, remove them by adjusting the pointers inward. Repeat this process until no more removals are possible, which occurs when the characters at both ends are distinct.

Efficient tracking with invariants

Keep track of the invariant condition that both ends of the string must match to remove characters. If the characters at both ends do not match, no further operations are possible. This invariant guides the loop condition and ensures that the time complexity remains linear, O(n).

Edge case handling

Handle cases where no characters can be removed. If the string is composed of different characters at both ends or consists entirely of one character, no operations can be performed, and the length of the string remains unchanged.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity is O(n), where n is the length of the string, because each character is checked once by the two-pointer technique. The space complexity is O(1) since we only need a few variables to track the pointers and the resulting string length.

What Interviewers Usually Probe

Check if the candidate efficiently uses the two-pointer approach to minimize the number of steps.
Look for correct identification of the invariant and its application in halting further removals.
Evaluate how well the candidate handles edge cases, such as strings that cannot be reduced further.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly manage the two pointers, leading to an incorrect stopping condition.
Overcomplicating the solution by using extra data structures when a simple two-pointer approach suffices.
Not handling edge cases, such as strings where no removal is possible or when all characters are the same.

Follow-up variants

Handling larger inputs efficiently, ensuring the solution works for strings with up to 100,000 characters.
Adapting the solution to allow for different operations at each end (e.g., only removing matching prefixes).
Modifying the problem to allow removals from non-matching parts of the string rather than only from the ends.

FAQ

How do I approach the Minimum Length of String After Deleting Similar Ends problem?

The problem can be efficiently solved using the two-pointer approach, where you compare the characters at both ends of the string and remove them if they are the same. This continues until no more operations are possible.

What is the time complexity of the solution?

The time complexity of the solution is O(n), where n is the length of the string. The two-pointer technique ensures that each character is checked once.

What are some common mistakes to avoid in this problem?

Avoid overcomplicating the solution with extra data structures or not managing the two pointers correctly. Also, ensure that edge cases are handled properly.

Can this problem be solved with a brute-force approach?

A brute-force approach could work but would be inefficient, as repeatedly removing characters from both ends could result in a time complexity worse than O(n). The two-pointer approach is optimal.

How does the two-pointer approach help in solving this problem?

The two-pointer approach helps efficiently check and remove characters from both ends of the string in linear time, reducing the problem to a manageable O(n) time complexity.

terminal

Solution

Solution 1: Two pointers

We define two pointers $i$ and $j$ to point to the head and tail of the string $s$ respectively, then move them to the middle until the characters pointed to by $i$ and $j$ are not equal, then $\max(0, j - i + 1)$ is the answer.

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class Solution:
    def minimumLength(self, s: str) -> int:
        i, j = 0, len(s) - 1
        while i < j and s[i] == s[j]:
            while i + 1 < j and s[i] == s[i + 1]:
                i += 1
            while i < j - 1 and s[j - 1] == s[j]:
                j -= 1
            i, j = i + 1, j - 1
        return max(0, j - i + 1)

Continue Practicing

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