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Piyush Acharya
Piyush Acharya

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SIMPLE & EFFICIENT Java Solution BEATS 95%

Intuition

To solve this problem, I need to keep track of the value of x and update it according to the operations in the array.

Approach

I will use a variable x to store the value of x and initialize it to zero. Then I will loop through the array of operations and use a switch statement to check each operation and increment or decrement x accordingly.

Complexity

  • Time complexity:
    The time complexity is $$O(n)$$ where n is the length of the array, because I need to iterate through all the operations once.

  • Space complexity:
    The space complexity is $$O(1)$$ because I only use a constant amount of extra space for the variable x.

Code

class Solution {
 public int finalValueAfterOperations(String[] operations) {
 int x = 0;
 int size = operations.length; // store the size of the array
 for (int i = 0; i < size; i++) { // use the size variable
 String operation = operations[i]; // store the operation in a variable
 switch (operation) { // use switch statement
 case "++X":
 case "X++":
 x++;
 break;
 case "--X":
 case "X--":
 x--;
 break;
 }
 }
 return x;
 }
}

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Top comments (0)

Great read:

Is it Time to go Back to the Monolith?

History repeats itself. Everything old is new again and I’ve been around long enough to see ideas discarded, rediscovered and return triumphantly to overtake the fad. In recent years SQL has made a tremendous comeback from the dead. We love relational databases all over again. I think the Monolith will have its space odyssey moment again. Microservices and serverless are trends pushed by the cloud vendors, designed to sell us more cloud computing resources.

Microservices make very little sense financially for most use cases. Yes, they can ramp down. But when they scale up, they pay the costs in dividends. The increased observability costs alone line the pockets of the “big cloud” vendors.