# Solving Leetcode Interviews in Seconds with AI: Final Value of Variable After Performing Operations


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```