algoadvance

Leetcode 2011. Final Value of Variable After Performing Operations

Problem Statement

You are given an integer variable x initialized to 0. You are also given a string array operations where operations[i] is one of "++X", "X++", "--X", and "X--". You need to perform the operations on x and return the final value of x.

Clarifying Questions

  1. Q: Are there any constraints on the length of the operations array? A: The length of the operations array will be within the range 1 to 100.

  2. Q: Are there any other types of operations except the four specified? A: No, the operations will strictly be one of "++X", "X++", "--X", and "X--".

  3. Q: Can the operations array contain duplicate operations? A: Yes, the array can contain duplicates of the specified operations.

Strategy

  1. Initialize a variable x to 0.
  2. Iterate through each operation in the operations array.
  3. For each operation:
    • Increment x if the operation is "++X" or "X++".
    • Decrement x if the operation is "--X" or "X--".
  4. Return the final value of x after processing all operations.

Code

public class Solution {
    public int finalValueAfterOperations(String[] operations) {
        int x = 0;
        for (String operation : operations) {
            if (operation.equals("++X") || operation.equals("X++")) {
                x++;
            } else if (operation.equals("--X") || operation.equals("X--")) {
                x--;
            }
        }
        return x;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String[] operations = {"++X", "X++", "--X", "--X"};
        System.out.println(solution.finalValueAfterOperations(operations)); // Output should be 0
    }
}

Time Complexity

The time complexity of this solution is O(n), where n is the length of the operations array. This is because we are iterating through the array once and performing constant time operations for each element.

Space Complexity

The space complexity is O(1), as we are only using a fixed amount of extra space regardless of the input size (int x).

This solution efficiently handles the problem within the provided constraints and ensures correct results by properly processing each element in the operations array.

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