Leetcode #2957: Remove Adjacent Almost-Equal Characters
In this guide, we solve Leetcode #2957 Remove Adjacent Almost-Equal Characters in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string word. In one operation, you can pick any index i of word and change word[i] to any lowercase English letter.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: word = "aaaaa"
Output: 2
Explanation: We can change word into "acaca" which does not have any adjacent almost-equal characters.
It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 2.
Python Solution
class Solution:
def removeAlmostEqualCharacters(self, word: str) -> int:
ans = 0
i, n = 1, len(word)
while i < n:
if abs(ord(word[i]) - ord(word[i - 1])) < 2:
ans += 1
i += 2
else:
i += 1
return ans
Complexity
The time complexity is , where is the length of the string word. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.