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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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