Formatted question description: https://leetcode.ca/all/2011.html

# 2011. Final Value of Variable After Performing Operations (Easy)

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--".

## Solution 1.

// OJ: https://leetcode.com/problems/final-value-of-variable-after-performing-operations/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int finalValueAfterOperations(vector<string>& A) {
int x = 0;
for (auto &s : A) {
if (s[0] == '+' || s[1] == '+') ++x;
else --x;
}
return x;
}
};