Fraction Addition and Subtraction

Fraction Addition and Subtraction - Problem

Given a string expression representing an expression of fraction addition and subtraction, return the calculation result in string format.

The final result should be an irreducible fraction. If your final result is an integer, change it to the format of a fraction that has a denominator 1. So in this case, 2 should be converted to 2/1.

Input & Output

Example 1 — Basic Mixed Operations

$ Input: expression = "-1/2+1/2+1/3"

› Output: "1/3"

💡 Note: First: -1/2 + 1/2 = 0, then 0 + 1/3 = 1/3. The result is already in irreducible form.

Example 2 — Multiple Fractions

$ Input: expression = "1/3-1/2"

› Output: "-1/6"

💡 Note: To subtract 1/3 - 1/2, we find common denominator: 2/6 - 3/6 = -1/6

Example 3 — Result is Integer

$ Input: expression = "1/2+1/2"

› Output: "1/1"

💡 Note: 1/2 + 1/2 = 1, which must be expressed as fraction 1/1

Constraints

The input string only contains '0' to '9', '/', '+' and '-'
Each fraction is in the form ±numerator/denominator
Each numerator and denominator is in the range [1, 10]
The final result should be an irreducible fraction

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to parse fractions sequentially and maintain a running sum using common denominator arithmetic. The single pass approach with character-by-character parsing is most efficient. Time: O(n), Space: O(1).

Common Approaches

✓ Binary Search

⏱️ Time: N/A Space: N/A

Parse and Calculate Step by Step

⏱️ Time: O(n) Space: O(1)

Extract each fraction and its sign from the expression, then process them one by one. For each operation, find the common denominator and perform the calculation.

Single Pass with Running Sum

⏱️ Time: O(n) Space: O(1)

Use a single pass to parse the expression character by character, maintaining a running sum of fractions. Apply GCD reduction only at the end for maximum efficiency.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int abs_val(int x) {
    return x < 0 ? -x : x;
}

int gcd(int a, int b) {
    a = abs_val(a);
    b = abs_val(b);
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

char* solution(char* expression) {
    int numerator = 0;
    int denominator = 1;
    
    int i = 0;
    int len = strlen(expression);
    
    while (i < len) {
        int sign = 1;
        if (expression[i] == '+' || expression[i] == '-') {
            if (expression[i] == '-') sign = -1;
            i++;
        } else if (i == 0 && expression[i] == '-') {
            sign = -1;
            i++;
        }
        
        int num = 0;
        while (i < len && expression[i] != '/') {
            num = num * 10 + (expression[i] - '0');
            i++;
        }
        i++; // skip '/'
        
        int den = 0;
        while (i < len && expression[i] != '+' && expression[i] != '-') {
            den = den * 10 + (expression[i] - '0');
            i++;
        }
        
        num *= sign;
        
        // Add fractions: numerator/denominator + num/den
        numerator = numerator * den + num * denominator;
        denominator = denominator * den;
        
        // Reduce fraction
        int g = gcd(numerator, denominator);
        numerator /= g;
        denominator /= g;
    }
    
    if (denominator < 0) {
        numerator = -numerator;
        denominator = -denominator;
    }
    
    char* result = malloc(50);
    sprintf(result, "%d/%d", numerator, denominator);
    return result;
}

int main() {
    char expression[1000];
    fgets(expression, sizeof(expression), stdin);
    
    // Remove newline and quotes
    int len = strlen(expression);
    if (expression[len-1] == '\n') expression[len-1] = '\0';
    
    if (expression[0] == '"') {
        memmove(expression, expression + 1, strlen(expression));
        len = strlen(expression);
        if (expression[len-1] == '"') expression[len-1] = '\0';
    }
    
    char* result = solution(expression);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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