Problem Description

You are given a simple programming language that has only four operations and one variable X:

• ++X and X++ both increment the value of variable X by 1
• --X and X-- both decrement the value of variable X by 1

The variable X starts with an initial value of 0.

You are given an array of strings called operations, where each string represents one of these four operations. Your task is to execute all the operations in order and return the final value of X.

For example:

• If operations = ["++X", "X++", "--X"], the value of X would change as follows:
  • Start: X = 0
  • After "++X": X = 1
  • After "X++": X = 2
  • After "--X": X = 1
  • Final result: 1

The key insight is that regardless of whether the ++ or -- appears before or after X, the effect on the variable is the same - increment by 1 for ++ operations and decrement by 1 for -- operations.

The solution cleverly identifies the operation type by checking the middle character (index 1) of each operation string. If it's '+', we add 1; if it's '-', we subtract 1. The sum of all these values gives us the final value of X.

Intuition

The first observation is that we need to track how the value of X changes after each operation. Since X starts at 0, we need to count the cumulative effect of all operations.

Looking at the four possible operations:

• ++X increments X by 1
• X++ increments X by 1
• --X decrements X by 1
• X-- decrements X by 1

We notice that whether the ++ or -- appears before or after X doesn't matter for the final value - both ++X and X++ have the same effect of adding 1, and both --X and X-- have the same effect of subtracting 1.

This means we don't need to distinguish between prefix and postfix operations. We only need to know whether each operation is an increment or decrement.

How can we quickly identify the operation type? If we examine the structure of each operation string:

• ++X has + at positions 0 and 1
• X++ has + at positions 1 and 2
• --X has - at positions 0 and 1
• X-- has - at positions 1 and 2

The character at index 1 (the middle position) always tells us the operation type! If operations[i][1] is '+', it's an increment operation; if it's '-', it's a decrement operation.

Therefore, we can simply iterate through all operations, check the character at index 1 of each string, and add 1 for '+' or subtract 1 for '-'. The sum of all these values gives us the final value of X.

Solution Approach

The solution uses a simple counting approach to calculate the final value of X.

Implementation Steps:

1. Traverse the operations array: We iterate through each operation string in the operations list.

2. Identify operation type: For each operation string s, we check the character at index 1 (the middle character):

   • If s[1] == '+', this is an increment operation, so we add 1
   • If s[1] == '-', this is a decrement operation, so we add -1

3. Calculate the sum: We use Python's sum() function with a generator expression to calculate the total effect of all operations in a single line.

The complete implementation:

def finalValueAfterOperations(self, operations: List[str]) -> int:
    return sum(1 if s[1] == '+' else -1 for s in operations)

This generator expression (1 if s[1] == '+' else -1 for s in operations) produces:

• 1 for each increment operation (++X or X++)
• -1 for each decrement operation (--X or X--)

The sum() function then adds all these values together, giving us the final value of X.

Time Complexity: O(n) where n is the number of operations, as we traverse the array once.

Space Complexity: O(1) as we only use a constant amount of extra space (the generator expression doesn't create an intermediate list).

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: operations = ["X++", "++X", "--X", "X--", "++X"]

Step-by-step execution:

1. Start with X = 0

2. Process each operation by examining the character at index 1:

   • "X++": Character at index 1 is '+' → Add 1 to our sum
   • "++X": Character at index 1 is '+' → Add 1 to our sum
   • "--X": Character at index 1 is '-' → Add -1 to our sum
   • "X--": Character at index 1 is '-' → Add -1 to our sum
   • "++X": Character at index 1 is '+' → Add 1 to our sum

3. Calculate the final value:

   • Sum = 1 + 1 + (-1) + (-1) + 1 = 1
   • Final value of X = 0 + 1 = 1

Visual representation of checking index 1:

"X++" → [X][+][+] → index 1 = '+' → contribute +1
"++X" → [+][+][X] → index 1 = '+' → contribute +1  
"--X" → [-][-][X] → index 1 = '-' → contribute -1
"X--" → [X][-][-] → index 1 = '-' → contribute -1
"++X" → [+][+][X] → index 1 = '+' → contribute +1

Using the one-line solution:

sum(1 if s[1] == '+' else -1 for s in ["X++", "++X", "--X", "X--", "++X"])
# Generates: 1, 1, -1, -1, 1
# Sum: 1 + 1 - 1 - 1 + 1 = 1

The final answer is 1, which matches what we'd get if we manually tracked X through each operation.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array operations. This is because the code iterates through each element in the operations list exactly once using a generator expression within the sum() function. For each operation, it performs a constant-time check (s[1] == '+') to determine whether to add 1 or -1.

The space complexity is O(1). The generator expression used in the sum() function doesn't create an intermediate list but instead generates values on-the-fly. The only additional space used is for the accumulator variable in the sum() function, which is constant regardless of the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Parse the Entire Operation String

A common mistake is trying to check for the exact operation format by comparing the entire string:

# Inefficient approach - checking entire strings
def finalValueAfterOperations(self, operations: List[str]) -> int:
    result = 0
    for op in operations:
        if op == "++X" or op == "X++":
            result += 1
        elif op == "--X" or op == "X--":
            result -= 1
    return result

Why it's problematic: This approach requires multiple string comparisons and is less elegant. It also becomes harder to maintain if the format changes slightly.

Solution: Check only the middle character (index 1) as shown in the optimal solution, since both ++ and -- operations have the operator symbol at index 1 regardless of whether it's prefix or postfix.

2. Index Out of Bounds Risk

If you don't validate the input, accessing operation[1] could raise an IndexError:

# Risky if operations contain empty or single-character strings
return sum(1 if s[1] == '+' else -1 for s in operations)

Solution: Add input validation if the problem doesn't guarantee valid inputs:

def finalValueAfterOperations(self, operations: List[str]) -> int:
    if not operations:
        return 0
  
    result = 0
    for op in operations:
        if len(op) >= 3:  # Ensure string has at least 3 characters
            result += 1 if op[1] == '+' else -1
    return result

3. Misunderstanding the Problem - Thinking Order Matters

Some might overthink and believe that ++X and X++ have different effects (like in C++ where prefix and postfix have different behaviors in expressions):

# Overcomplicated - treating prefix and postfix differently
def finalValueAfterOperations(self, operations: List[str]) -> int:
    result = 0
    for op in operations:
        if op == "++X":
            result += 1  # Increment before
        elif op == "X++":
            result += 1  # Increment after (same effect here!)
        # ... similar for decrement

Solution: Recognize that in this problem, both prefix and postfix operations have the same effect on the variable's final value. The position of the operator doesn't matter - only whether it's ++ or --.

4. Using Unnecessary Conditional Branches

Writing verbose if-elif-else chains when a simpler ternary operator suffices:

# Verbose approach
def finalValueAfterOperations(self, operations: List[str]) -> int:
    result = 0
    for op in operations:
        if op[1] == '+':
            result = result + 1
        else:
            result = result - 1
    return result

Solution: Use the concise generator expression with ternary operator as shown in the optimal solution, or at least use += and -= operators for cleaner code.