LeetCode 2011 - Final Value of Variable After Performing Operations

LeetCode Problem 2011

Difficulty: 🟢 Easy
Topics: Array, String, Simulation

Solution

Problem Understanding

This problem asks us to simulate a simple programming language with a single integer variable X that starts at 0. We are given a list of string operations, each of which either increments (++X or X++) or decrements (--X or X--) the value of X by 1. The goal is to determine the final value of X after performing all operations in the given order.

The input operations is an array of strings, where each string is guaranteed to be one of the four valid operations. The output is a single integer representing the final value of X. The constraints tell us that the array length ranges from 1 to 100, which is very small, so even a straightforward simulation will perform efficiently.

Edge cases to consider include an array of all increment operations, all decrement operations, alternating operations that cancel each other out, and the minimal array size of 1 operation. Since the input is guaranteed to contain only valid operations, we do not need to handle invalid strings or null input.

Approaches

The brute-force approach is straightforward: iterate through the array, examine each operation, and increment or decrement X accordingly. This works correctly because each operation affects X independently, and the operations are commutative with respect to each individual change. Given the constraints, this method is already fast, but it is technically O(n) where n is the number of operations.

The key observation for an optimal approach is that we do not need to distinguish between pre-increment (++X) and post-increment (X++) or pre-decrement (--X) and post-decrement (X--). Both forms result in the same change of +1 or -1 respectively. Therefore, we can treat any operation starting with + as +1 and any operation starting with - as -1. This simplifies the implementation and slightly reduces conditional logic, but it does not change the time complexity.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n)	O(1)	Iterate and apply each operation with conditionals
Optimal	O(n)	O(1)	Iterate and apply simplified logic based on first character

Algorithm Walkthrough

1. Initialize a variable X to 0. This represents the starting value of the variable.
2. Iterate through each string in the operations array.
3. For each operation, check the first character of the string. If it is '+', increment X by 1. If it is '-', decrement X by 1.
4. Continue until all operations have been processed.
5. Return the final value of X.

Why it works: The algorithm works because each operation has an isolated effect on X and there is no dependency between operations other than sequential application. Treating the first character as the indicator of increment or decrement is valid since all operations are guaranteed to be one of the four specified forms. The final result reflects the cumulative effect of all operations.

Python Solution

from typing import List

class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        X = 0
        for op in operations:
            if op[0] == '+':
                X += 1
            else:
                X -= 1
        return X

The Python implementation initializes X to 0 and iterates through each operation. The first character of each string is checked to determine whether to increment or decrement X. This matches the algorithm walkthrough exactly and guarantees the correct final value.

Go Solution

func finalValueAfterOperations(operations []string) int {
    X := 0
    for _, op := range operations {
        if op[0] == '+' {
            X += 1
        } else {
            X -= 1
        }
    }
    return X
}

In Go, the solution is almost identical. We iterate through the slice of strings using a range loop. The first byte of each string is checked (op[0]) to determine increment or decrement. Since the input is guaranteed to be valid, we do not need to check the string length or handle errors.

Worked Examples

Example 1: ["--X","X++","X++"]

Operation	X before	X after
--X	0	-1
X++	-1	0
X++	0	1

Final value: 1

Example 2: ["++X","++X","X++"]

Operation	X before	X after
++X	0	1
++X	1	2
X++	2	3

Final value: 3

Example 3: ["X++","++X","--X","X--"]

Operation	X before	X after
X++	0	1
++X	1	2
--X	2	1
X--	1	0

Final value: 0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We process each operation once
Space	O(1)	Only a single integer variable is used

The reasoning is straightforward: we iterate through the operations array once, performing a constant-time update for each operation, and we store no additional data structures.

Test Cases

# Provided examples
assert Solution().finalValueAfterOperations(["--X","X++","X++"]) == 1  # Example 1
assert Solution().finalValueAfterOperations(["++X","++X","X++"]) == 3  # Example 2
assert Solution().finalValueAfterOperations(["X++","++X","--X","X--"]) == 0  # Example 3

# Edge cases
assert Solution().finalValueAfterOperations(["X++"]) == 1  # Single increment
assert Solution().finalValueAfterOperations(["X--"]) == -1  # Single decrement
assert Solution().finalValueAfterOperations(["++X","--X"]) == 0  # Canceling operations
assert Solution().finalValueAfterOperations(["X++"]*100) == 100  # Max size array, all increments
assert Solution().finalValueAfterOperations(["X--"]*100) == -100  # Max size array, all decrements

Test	Why
["--X","X++","X++"]	Mixed operations leading to positive final value
["++X","++X","X++"]	All increment operations
["X++","++X","--X","X--"]	Alternating operations canceling to zero
["X++"]	Minimal input size
["X--"]	Minimal input size with decrement
["++X","--X"]	Operations that cancel
["X++"]*100	Maximum array size with all increments
["X--"]*100	Maximum array size with all decrements

Edge Cases

One important edge case is when the operations array contains only one element. This tests whether the implementation handles minimal input correctly and returns the correct increment or decrement.

Another edge case is when operations cancel each other out, such as ["++X","--X","X++","X--"]. This ensures the algorithm correctly applies each operation sequentially without accumulating errors.

Finally, the maximal array size of 100 operations is an edge case for performance. While the algorithm handles it trivially due to O(n) time complexity, it is still important to verify that it performs correctly under repeated identical operations, especially if all operations are increments or all are decrements. This confirms there are no integer overflow issues, which in Python is not a concern but in some languages might require attention.