How do you solve Remove Adjacent Almost-Equal Characters on LeetCode?

Remove Adjacent Almost-Equal Characters (LeetCode #2957) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Adjacent characters can be efficiently handled by ensuring we change them to a character that is not adjacent to either neighbor.

What is the time complexity of Remove Adjacent Almost-Equal Characters?

The optimal solution for Remove Adjacent Almost-Equal Characters runs in O(n) time and uses O(1) space. The complexity is O(n) because we only make a single pass through the string, checking each character against its predecessor.

What is the brute force solution for Remove Adjacent Almost-Equal Characters?

The brute force approach for Remove Adjacent Almost-Equal Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each character in the string and counting how many changes are needed to ensure no two adjacent characters are almost-equal. 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 Remove Adjacent Almost-Equal Characters?

Remove Adjacent Almost-Equal Characters uses the following concepts: String, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Remove Adjacent Almost-Equal Characters?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can minimize operations — sometimes a greedy approach is the best. Practice dry-running your code with different examples to ensure it handles all cases.

What are common mistakes when solving Remove Adjacent Almost-Equal Characters?

Failing to consider edge cases where the first or last character is involved. Not realizing that multiple operations can be required for consecutive characters.

#2957

Remove Adjacent Almost-Equal Characters

Medium

StringDynamic ProgrammingGreedyGreedyDynamic Programming

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 greedy approach where we iterate through the string and change characters only when necessary. This ensures we minimize the number of operations while maintaining the constraints.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for operations needed.
2Step 2: Iterate through the string from the first character to the last.
3Step 3: For each character, check if it is almost-equal to the previous character. If it is, increment the operation counter and change the character to a valid non-adjacent character.

solution.py12 lines

1def min_operations(word):
2    operations = 0
3    n = len(word)
4    for i in range(1, n):
5        if abs(ord(word[i]) - ord(word[i - 1])) <= 1:
6            operations += 1
7            # Change to a non-adjacent character
8            if word[i - 1] == 'z':
9                word = word[:i] + 'x' + word[i + 1:]
10            else:
11                word = word[:i] + chr(ord(word[i - 1]) + 2) + word[i + 1:]
12    return operations

ℹ

Complexity note: The complexity is O(n) because we only make a single pass through the string, checking each character against its predecessor.

1Adjacent characters can be efficiently handled by ensuring we change them to a character that is not adjacent to either neighbor.
2The greedy approach minimizes changes by making local optimal choices.

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