Introduction

In this blog post, we will explore how to solve the LeetCode problem "2011" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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--".

Explanation

• Iterate through the operations: The solution iterates through the input array of operations.
  • Increment or decrement: For each operation, it checks if it's an increment or decrement operation. Based on the operation, the value of X is updated accordingly.
  • Return the final value: After processing all the operations, the final value of X is returned.

• Runtime Complexity: O(n), where n is the number of operations. Storage Complexity: O(1).

Code

    class Solution:
    def finalValueAfterOperations(self, operations: list[str]) -> int:
        x = 0
        for operation in operations:
            if operation == "++X" or operation == "X++":
                x += 1
            elif operation == "--X" or operation == "X--":
                x -= 1
        return x