2011. Final Value of Variable After Performing Operations LeetCode Solution
In this guide, you will get 2011. Final Value of Variable After Performing Operations LeetCode Solution with the best time and space complexity. The solution to Final Value of Variable After Performing Operations problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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.
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–".
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
2011. Final Value of Variable After Performing Operations LeetCode Solution in C++
class Solution {
public:
int finalValueAfterOperations(vector<string>& operations) {
int ans = 0;
for (const string& op : operations)
ans += op[1] == '+' ? 1 : -1;
return ans;
}
}; /* code provided by PROGIEZ */
2011. Final Value of Variable After Performing Operations LeetCode Solution in Java
class Solution {
public int finalValueAfterOperations(String[] operations) {
int ans = 0;
for (final String op : operations)
ans += op.charAt(1) == '+' ? 1 : -1;
return ans;
}
} // code provided by PROGIEZ
2011. Final Value of Variable After Performing Operations LeetCode Solution in Python
class Solution:
def finalValueAfterOperations(self, operations: list[str]) -> int:
return sum(op[1] == '+' or -1 for op in operations) # code by PROGIEZ
