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2011. Final Value of Variable After Performing Operations

Description

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--".

Solutions

Solution 1: Simulation

Traverse the array operations. For each operation $operations[i]$, if it contains '+', then the answer increases by $1$, otherwise the answer decreases by $1$.

The time complexity is $O(n)$, where $n$ is the length of the array operations. The space complexity is $O(1)$.

  • class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int ans = 0;
        for (var s : operations) {
            ans += (s.charAt(1) == '+' ? 1 : -1);
        }
        return ans;
    }
}

  • class Solution {
public:
    int finalValueAfterOperations(vector<string>& operations) {
        int ans = 0;
        for (auto& s : operations) ans += (s[1] == '+' ? 1 : -1);
        return ans;
    }
};

  • class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        return sum(1 if s[1] == '+' else -1 for s in operations)


  • func finalValueAfterOperations(operations []string) (ans int) {
	for _, s := range operations {
		if s[1] == '+' {
			ans += 1
		} else {
			ans -= 1
		}
	}
	return
}

  • function finalValueAfterOperations(operations: string[]): number {
    let ans = 0;
    for (let operation of operations) {
        ans += operation.includes('+') ? 1 : -1;
    }
    return ans;
}


  • /**
 * @param {string[]} operations
 * @return {number}
 */
var finalValueAfterOperations = function (operations) {
    let ans = 0;
    for (const s of operations) {
        ans += s[1] === '+' ? 1 : -1;
    }
    return ans;
};


  • impl Solution {
    pub fn final_value_after_operations(operations: Vec<String>) -> i32 {
        let mut ans = 0;
        for s in operations.iter() {
            ans += if s.as_bytes()[1] == b'+' { 1 } else { -1 };
        }
        ans
    }
}

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