#2011

Easy

LeetCode Problem Workspace

Final Value of Variable After Performing Operations

Compute the final value of X by simulating each string operation in the array sequentially with careful tracking.

Problem Statement

You are given an array of strings operations, each representing a simple operation on a single integer variable X, which starts at 0. Each string is either '++X', 'X++', '--X', or 'X--', and modifies X by incrementing or decrementing it by 1. Compute the final value of X after performing all operations in order.

Constraints: 1 <= operations.length <= 100, and each element of operations is guaranteed to be one of '++X', 'X++', '--X', or 'X--'. Return the integer representing X's value after completing all operations.

Examples

Example 1

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

Output: 1

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

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

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--".

Solution Approach

Iterate and Apply Operations

Loop through the operations array and for each string, check if it contains '++' or '--'. Increment X if '++' appears, decrement if '--' appears. This direct simulation ensures that the final X reflects all operations correctly.

Count Instead of Simulate

For optimization, count the number of increments and decrements separately by scanning the array once. Then compute final X as increments minus decrements. This reduces intermediate variable changes while preserving the correct final value.

One-Pass Inline Update

Maintain X as you iterate, updating it immediately for each operation string. This keeps memory usage minimal and avoids storing extra counters, aligning with the Array plus String pattern where string parsing drives state changes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since we process each operation once. Space complexity is O(1) because we only maintain the variable X and no additional structures.

What Interviewers Usually Probe

Check if candidates properly handle both pre- and post-increment forms equivalently.
Watch for solutions that overcomplicate with extra arrays or maps instead of simple counters.
Notice if they iterate multiple times unnecessarily rather than a single linear pass.

Common Pitfalls or Variants

Common pitfalls

Miscounting X when confusing pre-increment (++X) and post-increment (X++) forms.
Using extra data structures which are unnecessary and increase space complexity.
Failing to process the operations in order, which can lead to incorrect final X.

Follow-up variants

Operations could include different variable names, requiring mapping from names to values.
Allowing multi-step operations like 'X+=2' or 'X-=3', extending the simulation approach.
Instead of returning final X, return an array of X after each operation for step-by-step tracking.

FAQ

What is the main approach for Final Value of Variable After Performing Operations?

Iterate through the operations array, applying increments or decrements to X based on the string content, ensuring all operations are executed sequentially.

Does it matter if the operation is '++X' or 'X++'?

No, for the final value of X, both forms have the same net effect, so either increments X by 1 immediately.

Can we optimize without simulating each operation?

Yes, you can count total increments and decrements in a single pass and compute final X as the difference.

What is the time complexity of this solution?

The solution runs in O(n) time where n is the length of the operations array, since each operation is processed once.

How does the Array plus String pattern affect this problem?

The pattern shows that the array drives processing while string content determines the update, making string parsing central to computing X.

terminal

Solution

Solution 1: Counting

We traverse the array $\textit{operations}$. For each operation $\textit{operations}[i]$, if it contains `'+'`, we increment the answer by $1$, otherwise, we decrement the answer by $1$.

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

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