[파이썬][Leetcode][릿코드] Final Value of Variable After Performing Operations

1. 문제

2011. Final Value of Variable After Performing Operations

Easy

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There is a programming language with only four operations and one variable X:

• ++X and X++ increments the value of the variable X by 1.
• --X and X-- decrements the value of the variable X by 1.

Initially, the value of X is 0.

Given an array of strings operations containing a list of operations, return the final value of X after performing all the operations.

 

Example 1:

Input: operations = ["--X","X++","X++"]
Output: 1
Explanation: The operations are performed as follows:
Initially, X = 0.
--X: X is decremented by 1, X =  0 - 1 = -1.
X++: X is incremented by 1, X = -1 + 1 =  0.
X++: X is incremented by 1, X =  0 + 1 =  1.

Example 2:

Input: operations = ["++X","++X","X++"]
Output: 3
Explanation: The operations are performed as follows:
Initially, X = 0.
++X: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
X++: X is incremented by 1, X = 2 + 1 = 3.

Example 3:

Input: operations = ["X++","++X","--X","X--"]
Output: 0
Explanation: The operations are performed as follows:
Initially, X = 0.
X++: X is incremented by 1, X = 0 + 1 = 1.
++X: X is incremented by 1, X = 1 + 1 = 2.
--X: X is decremented by 1, X = 2 - 1 = 1.
X--: X is decremented by 1, X = 1 - 1 = 0.

 

Constraints:

• 1 <= operations.length <= 100
• operations[i] will be either "++X", "X++", "--X", or "X--".

출처: Leetcode, https://leetcode.com/

2. 해결방법 시간복잡도

1. 단순 코딩 O(N)
2. 더 쉽게 O(1)

3. 문제 해결 및 코드

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• 주석을 참고하면서 이해를 돕습니다.

4. 알고리즘 및 해설

1. 해당 문제에서 제시한 특정 문자에 대한 답으로 딕셔너리를 만든다.
2. 이후 해당 값이 키값에 존재한다면 해당 키에 값을 추가한다.
3. 이후 해당 키값들의 합을 출력한다.

5. 더 쉽게 알아보기

1. 그냥 count()로 쉽게 해결이 가능했다.

   class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        return operations.count("X++") + operations.count("++X") - operations.count("X--") - operations.count("--X")