Final Value of Variable After Performing Operations

Final Value of Variable After Performing Operations - Problem

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.

Input & Output

Example 1 — Mixed Operations

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

› Output: 2

💡 Note: X starts at 0. ++X: X=1, ++X: X=2, --X: X=1, X++: X=2. Final value is 2.

Example 2 — Only Increments

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

› Output: 2

💡 Note: X starts at 0. X++: X=1, ++X: X=2. Final value is 2.

Example 3 — Only Decrements

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

› Output: -2

💡 Note: X starts at 0. --X: X=-1, X--: X=-2. Final value is -2.

Constraints

1 ≤ operations.length ≤ 100
operations[i] will be either "++X", "X++", "--X", or "X--".

Visualization

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Understanding the Visualization

Input

Array of operation strings

Process

Execute each operation on variable X

Output

Final value of X

Key Takeaway

🎯 Key Insight: All increment/decrement operations have the same effect regardless of pre/post notation

Asked in

a Amazon 15 M Microsoft 12 G Google 8 f Facebook 5

The key insight is that all four operations (++X, X++, --X, X--) have the same effect - they either increment or decrement X by 1. Best approach is to simply check for '+' or '-' characters in each operation string. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized - Character Detection	O(n)	O(1)	Check only the first or second character to determine operation type
Brute Force - Check Each Operation	O(n)	O(1)	Iterate through each operation and update X accordingly

Optimized - Character Detection — Algorithm Steps

Initialize X = 0
For each operation, check if it contains '+' character
If '+' found, increment; otherwise decrement
Return final X

Visualization

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Step-by-Step Walkthrough

Initialize

Start with X = 0

Character Check

Look for '+' or '-' in each operation

Result

Return final value of X

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char operations[][10], int n) {
    int x = 0;
    for (int i = 0; i < n; i++) {
        if (strchr(operations[i], '+') != NULL) {
            x += 1;
        } else {
            x -= 1;
        }
    }
    return x;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char operations[100][10];
    int n = 0;
    
    char* token = strtok(line, "[],\"\n ");
    while (token != NULL) {
        if (strlen(token) > 0) {
            strcpy(operations[n], token);
            n++;
        }
        token = strtok(NULL, "[],\"\n ");
    }
    
    int result = solution(operations, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through n operations with O(1) character check

✓ Linear Growth

Space Complexity

O(1)

Only using one variable to store X

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized - Character Detection — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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