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Leetcode 2011. Final Value of Variable After Performing Operations

Problem Statement

Given an array of strings operations containing a list of operations on a variable X that start at 0, determine the final value of X after performing all the operations. Each operation is a string that can be one of the following: "--X", "X--", "++X", "X++".

• "++X" and "X++" increments the value of X by 1.
• "--X" and "X--" decrements the value of X by 1.

Clarifying Questions

1. What will be the maximum length of the operations array?
   • Typical size constraints apply, such as (1 \leq \text{operations.length} \leq 1000).
2. Are the operation strings always valid strings as described?
   • Yes, all strings in the operations array will be one of the four valid operations.
3. What should be the initial value of X?
   • The initial value of X is always 0.

Strategy

1. Initialize a variable X to 0.
2. Traverse through each string in the operations array.
3. For each operation, update the value of X accordingly:
   • 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 applying all operations.

Code

Here’s the implementation of the plan in C++:

#include <vector>
#include <string>

int finalValueAfterOperations(std::vector<std::string>& operations) {
    int X = 0;
    for (const std::string& operation : operations) {
        if (operation == "++X" || operation == "X++") {
            X++;
        } else if (operation == "--X" || operation == "X--") {
            X--;
        }
    }
    return X;
}

Explanation

• We start by initializing X as 0.
• We iterate over each string in the operations vector.
• Depending on the operation string, we modify the value of X:
  • If the operation string contains "++X" or "X++", increment X by 1.
  • If the operation string contains "--X" or "X--", decrement X by 1.
• Finally, return the value of X.

Time Complexity

• The time complexity of this solution is (O(n)), where (n) is the size of the operations array.
• This is because we are making a single pass through the operations array and each operation (increment/decrement) is performed in constant time.

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