Final Value of Variable After Performing Operations

LeetCode 2137 | Difficulty: Easy

Easy

Problem Description

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 <= 100

• operations[i] will be either "++X", "X++", "--X", or "X--".

Topics: Array, String, Simulation

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Approach

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

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Solutions

Solution 1: C# (Best: 125 ms)

Metric	Value
Runtime	125 ms
Memory	26.5 MB
Date	2021-09-20

Solution

public class Solution {
    public int FinalValueAfterOperations(string[] operations) {
        if (operations.Length == 0) { return 0;}	
	int additions = 0;
	int subtractions = 0;
	for (int i = 0; i < operations.Length; i++)
	{
		switch (operations[i])
		{
			case "--X":
			case "X--":
				subtractions++;
				break;
			case "X++":
			case "++X":
				additions++;
				break;
		}
	}
	return additions-subtractions;
    }
}

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• LeetCode provides 2 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: There are only two operations to keep track of.

Hint 2: Use a variable to store the value after each operation.