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.

Examples

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--".

Approaches

1️⃣ Iteration

Code:

 int finalValueAfterOperations(vector<string>& operations) {
        int value = 0;
        for(int i = 0; i < operations.size(); i++){
            string operation = operations[i];
            
            if (operation == "--X" || operation == "X--"){
                    value--;
            } else if(operation == "++X" || operation == "X++"){
                    value++;
            }
        }
        return value;
        
    }

Complexity Analysis:

• Time complexity: O(N).
• Space complexity: O(1).