Equalize Strings by Adding or Removing Characters at Ends Python/Java solution

Implementations typically use dynamic programming to compute the longest common substring. The DP table compares characters from both strings and extends matches diagonally. FleetCode provides working implementations in Python, Java, C++, Go, and TypeScript following this approach.

What is the best approach for Equalize Strings by Adding or Removing Characters at Ends?

The most reliable approach is computing the longest common substring using dynamic programming. Once the longest shared contiguous substring of length L is found, the minimum operations required is (n − L) + (m − L). This approach runs in O(n * m) time and directly models the constraint that characters can only be removed from the ends.

Is Equalize Strings by Adding or Removing Characters at Ends asked at Google/Amazon/Meta?

Problems involving longest common substring, substring hashing, and dynamic programming frequently appear in interviews at companies like Google, Amazon, and Meta. This problem combines those patterns with string manipulation constraints, making it a realistic medium-level interview question.

What data structure is used in Equalize Strings by Adding or Removing Characters at Ends?

The primary structure is a 2D dynamic programming table for computing the longest common substring. Optimized approaches also use hash sets for substring hashes, rolling hash arrays, and sliding window techniques to check substring matches efficiently.

What is the time complexity of Equalize Strings by Adding or Removing Characters at Ends?

The standard dynamic programming solution runs in O(n * m) time and O(n * m) space, where n and m are the lengths of the two strings. An optimized version using binary search with rolling hash can reduce the time complexity to O((n + m) log(min(n, m))).

3135. Equalize Strings by Adding or Removing Characters at Ends - Practice Online

Q: Is Equalize Strings by Adding or Removing Characters at Ends easy or hard?

This problem is generally classified as Medium difficulty. The main challenge is recognizing that removing characters from the ends preserves a contiguous substring, which reduces the problem to finding the longest common substring between the two strings.

Q: How to solve Equalize Strings by Adding or Removing Characters at Ends in O(n)?

A strict O(n) solution generally isn't achievable for arbitrary strings because the problem requires comparing substrings across both inputs. The closest efficient approach is binary search combined with rolling hash, which runs in O((n + m) log(min(n, m))) time while checking substring matches efficiently.

Given two strings initial and target, your task is to modify initial by performing a series of operations to make it equal to target.

In one operation, you can add or remove one character only at the beginning or the end of the string initial.

Return the minimum number of operations required to transform initial into target.

Example 1:

Input: initial = "abcde", target = "cdef"

Output: 3

Explanation:

Remove 'a' and 'b' from the beginning of initial, then add 'f' to the end.

Example 2:

Input: initial = "axxy", target = "yabx"

Output: 6

Explanation:

Operation	Resulting String
Add `'y'` to the beginning	`"yaxxy"`
Remove from end	`"yaxx"`
Remove from end	`"yax"`
Remove from end	`"ya"`
Add `'b'` to the end	`"yab"`
Add `'x'` to the end	`"yabx"`

Example 3:

Input: initial = "xyz", target = "xyz"

Output: 0

Explanation:

No operations are needed as the strings are already equal.

Constraints:

1 <= initial.length, target.length <= 1000
initial and target consist only of lowercase English letters.