Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 20

Description

Final Value of Variable After Performing Operations

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

class Solution {
public:
  int finalValueAfterOperations(vector<string>& operations) {
    int value = 0;
    for(const auto &op : operations) {
      if(op[1] == '+') {
        value += 1;
      } else {
        value -= 1;
      }
    }
    return value;
  }
};

// Accepted
// 259/259 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 42.92 % of cpp submissions (17.7 MB)