What is the brute force solution for Final Value of Variable After Performing Operations?

The brute force approach for Final Value of Variable After Performing Operations is Brute Force, with O(n) time complexity and O(1) space complexity. We can solve this problem by iterating through each operation and updating the value of X accordingly. This is straightforward since we just need to apply each operation one by one. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Final Value of Variable After Performing Operations?

Final Value of Variable After Performing Operations uses the following concepts: Array, String, Simulation. The recommended patterns to study are: Simulation, Counting.

What are the interview tips for Final Value of Variable After Performing Operations?

Always start with a clear understanding of the problem and the initial conditions. Think about how you can simplify the problem; counting operations can often reduce complexity. Be mindful of edge cases, such as all operations being increments or decrements.

What are common mistakes when solving Final Value of Variable After Performing Operations?

Confusing the effects of pre-increment and post-increment; they both result in the same net change in this context. Not considering the initial value of X when performing operations.

#2011

Final Value of Variable After Performing Operations

Easy

ArrayStringSimulationSimulationCounting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of simulating each operation, we can count the number of increments and decrements directly. This allows us to compute the final value of X in a single pass.

⚙️

Algorithm

4 steps

1Step 1: Initialize X to 0.
2Step 2: Loop through each operation in the operations array.
3Step 3: For each operation, if it is an increment, increase a counter; if it is a decrement, decrease a counter.
4Step 4: Return the final value of X as the difference between increments and decrements.

solution.py2 lines

1def finalValueAfterOperations(operations):
2    return sum(1 if op in ['++X', 'X++'] else -1 for op in operations)

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the operations array once. The space complexity is O(1) because we only use a fixed amount of extra space for the count variable.

1The operations can be categorized into increments and decrements, allowing for a more efficient counting approach.
2The order of operations does not affect the final result, only the total counts of increments and decrements matter.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.