There is a programming language with only four operations and one variable X:
++X and X++ increments the value of the variable X by 1.--X and X-- decrements the value of the variable X by 1.Initially, the value of X is 0.
Given an array of strings operations containing a list of operations, return the final value of X after performing all the operations.
Example 1:
Input: operations = ["--X","X++","X++"] Output: 1 Explanation: 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 Explanation: 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 Explanation: 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 <= 100operations[i] will be either "++X", "X++", "--X", or "X--".When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
We start with a variable and a list of actions to perform on it. The brute-force way is to go through the list of actions one by one and apply each action to the variable in the order they appear.
Here's how the algorithm would work step-by-step:
def final_value_after_operations_brute_force(operations):
variable_value = 0
# Iterate through the list of operations.
for operation in operations:
# Check if the operation is incrementing or decrementing.
if operation == "--X" or operation == "X--":
# Apply the decrement to the variable.
variable_value -= 1
elif operation == "++X" or operation == "X++":
# Apply the increment to the variable.
variable_value += 1
return variable_valueThe key idea is to keep track of the current value of the variable. Since we only need the final result, we can process each operation one at a time, immediately updating the variable's value based on whether the operation increases or decreases it.
Here's how the algorithm would work step-by-step:
def final_value_after_operations(operations):
variable_value = 0
# Iterate through each operation.
for operation in operations:
# Check for increment operations.
if "++" in operation:
variable_value += 1
# Otherwise, it must be a decrement operation.
else:
variable_value -= 1
# Return the final value of the variable.
return variable_value| Case | How to Handle |
|---|---|
| Null or empty operations array | Return 0 if operations is null or empty as there are no operations to perform. |
| Operations array contains invalid operations (not ++X, X++, --X, X--) | Throw an IllegalArgumentException or return an error value to indicate invalid input. |
| Very large number of operations (approaching memory limits) | The solution should still perform correctly as each operation is processed independently, but memory usage should be monitored for extremely large inputs. |
| Integer overflow during increment/decrement | The problem statement does not mention overflow, but consider using a larger data type (long) or modulo arithmetic if the problem statement explicitly mentions possible overflows. |
| Operations with leading or trailing whitespace | Trim whitespace from each operation string before processing it. |
| Operations are a mix of increment and decrement, resulting in a final value of zero | The solution should correctly handle cancellations resulting in 0. |
| Operations array contains only increment or only decrement operations | The solution should perform as expected even when operations are homogeneous. |
| Read-only input operations array | The solution only reads from the input array, so it handles read-only arrays without issue. |