Final Value of Variable After Performing Operations
(LeetCode easy problem)
There is a programming language with only four operations and one variable X
:
++X
andX++
increments the value of the variableX
by1
.--X
andX--
decrements the value of the variableX
by1
.
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--"
.
The approach I have used here is the simplest, in case I encounter “++X” or “X++” I will simply increment the value of x, if I get, or "X--"
I will simply decrement the value. In case I encounter anything else I will simply continue.
The code is given below :
var finalValueAfterOperations = function(operations) {
let num = 0;
operations.forEach((operation, i)=>{
num = operation === '--X' || operation === 'X--' ? num - 1 : operation === 'X++' || operation === '++X' ? num + 1 : 0;
})
return num;
};
Time Complexity: O(n)
Space Complexity: O(1)
Runtime And Memory Usage
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