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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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

✓ Brute Force - Check Each Operation

⏱️ Time: O(n) Space: O(1)

Check each operation string individually using if-else conditions to determine whether to increment or decrement X. Start with X = 0 and process operations one by one.

Optimized - Character Detection

⏱️ Time: O(n) Space: O(1)

Instead of comparing entire strings, just check if the operation contains '+' or '-' character. This reduces the number of comparisons needed for each operation.

Brute Force - Check Each Operation — Algorithm Steps

Initialize X = 0
For each operation, check if it contains '+' or '-'
If '+', increment X; if '-', decrement X
Return final value of X

Visualization

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Step-by-Step Walkthrough

Initialize

Start with X = 0

Process Operations

Check each operation and update X

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 (strcmp(operations[i], "++X") == 0 || strcmp(operations[i], "X++") == 0) {
            x += 1;
        } else if (strcmp(operations[i], "--X") == 0 || strcmp(operations[i], "X--") == 0) {
            x -= 1;
        }
    }
    return x;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char operations[100][10];
    int n = 0;
    
    if (strlen(line) > 3) {
        char* token = strtok(line, "[],\"\n ");
        while (token != NULL) {
            if (strlen(token) > 0 && (strcmp(token, "++X") == 0 || strcmp(token, "X++") == 0 || strcmp(token, "--X") == 0 || strcmp(token, "X--") == 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

✓ Linear Growth

Space Complexity

O(1)

Only using one variable to store X

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Each Operation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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