Simplified Fractions

Simplified Fractions - Problem

Given an integer n, return a list of all simplified fractions between 0 and 1 (exclusive) such that the denominator is less-than-or-equal-to n.

A fraction is simplified if the greatest common divisor (GCD) of the numerator and denominator is 1.

You can return the answer in any order.

Input & Output

Example 1 — Basic Case

$ Input: n = 2

› Output: ["1/2"]

💡 Note: Only one fraction with denominator ≤ 2: 1/2 is already simplified since gcd(1,2) = 1

Example 2 — Multiple Denominators

$ Input: n = 3

› Output: ["1/2", "1/3", "2/3"]

💡 Note: Denominators 2 and 3: 1/2 (gcd=1), 1/3 (gcd=1), 2/3 (gcd=1). All are simplified.

Example 3 — With Non-Simplified

$ Input: n = 4

› Output: ["1/2", "1/3", "1/4", "2/3", "3/4"]

💡 Note: Skip 2/4 because gcd(2,4) = 2, so it simplifies to 1/2 which we already have.

Constraints

1 ≤ n ≤ 100

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is that a fraction is simplified when gcd(numerator, denominator) = 1. We generate all possible fractions and filter using GCD calculation. Best approach uses efficient Euclidean algorithm. Time: O(n² log n), Space: O(1)

Common Approaches

✓ Brute Force with GCD Check

⏱️ Time: O(n² log n) Space: O(1)

Generate all possible fractions with denominators from 2 to n and numerators from 1 to denominator-1. For each fraction, calculate GCD and keep only those where GCD equals 1.

Euler's Totient Function

⏱️ Time: O(n² log n) Space: O(1)

Instead of checking GCD for every fraction, we can use mathematical insights. A fraction a/b is simplified if gcd(a,b) = 1. We can optimize by avoiding redundant GCD calculations through better iteration patterns.

Brute Force with GCD Check — Algorithm Steps

Iterate through all denominators from 2 to n
For each denominator, try numerators from 1 to denominator-1
Calculate GCD of numerator and denominator
If GCD is 1, add fraction to result

Visualization

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Step-by-Step Walkthrough

Generate All

Create all fractions with denominators 2 to n

Check GCD

Calculate GCD for each fraction

Keep Simplified

Only keep fractions where GCD = 1

Code -

solution.c — C

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

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

void solution(int n) {
    printf("[");
    int first = 1;
    for (int denominator = 2; denominator <= n; denominator++) {
        for (int numerator = 1; numerator < denominator; numerator++) {
            if (gcd(numerator, denominator) == 1) {
                if (!first) printf(",");
                printf("\"%d/%d\"", numerator, denominator);
                first = 0;
            }
        }
    }
    printf("]\n");
}

int main() {
    int n;
    scanf("%d", &n);
    solution(n);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² log n)

Nested loops O(n²) with GCD calculation O(log n) for each fraction

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for computation, result space not counted

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force with GCD Check — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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