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Simplified Fractions

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

Given an integer n, return _a list of allsimplified fractions between _0 and1 (exclusive) such that the denominator is less-than-or-equal-ton. You can return the answer in any order.

Examples

Example 1

Input: n = 2
Output: ["1/2"]
Explanation: "1/2" is the only unique fraction with a denominator less-than-or-equal-to 2.

Example 2

Input: n = 3
Output: ["1/2","1/3","2/3"]

Example 3

Input: n = 4
Output: ["1/2","1/3","1/4","2/3","3/4"]
Explanation: "2/4" is not a simplified fraction because it can be simplified to "1/2".

Constraints

  • 1 <= n <= 100

Solution

Method 1 – Brute Force with GCD

Intuition

For each denominator d from 2 to n, and numerator num from 1 to d-1, if gcd(num, d) == 1, the fraction is simplified.

Approach

  1. For each denominator d from 2 to n:
    • For each numerator num from 1 to d-1:
      • If gcd(num, d) == 1, add "num/d" to the result.

Code

C++
#include <vector>
#include <string>
#include <numeric>
using namespace std;
class Solution {
public:
    vector<string> simplifiedFractions(int n) {
        vector<string> res;
        for (int d = 2; d <= n; ++d) {
            for (int num = 1; num < d; ++num) {
                if (gcd(num, d) == 1)
                    res.push_back(to_string(num) + "/" + to_string(d));
            }
        }
        return res;
    }
};
Java
import java.util.*;
class Solution {
    public List<String> simplifiedFractions(int n) {
        List<String> res = new ArrayList<>();
        for (int d = 2; d <= n; ++d) {
            for (int num = 1; num < d; ++num) {
                if (gcd(num, d) == 1)
                    res.add(num + "/" + d);
            }
        }
        return res;
    }
    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}
Python
from math import gcd
class Solution:
    def simplifiedFractions(self, n):
        res = []
        for d in range(2, n+1):
            for num in range(1, d):
                if gcd(num, d) == 1:
                    res.append(f"{num}/{d}")
        return res

Complexity

  • ⏰ Time complexity: O(n^2) — For each denominator, check all numerators.
  • 🧺 Space complexity: O(n^2) — For the result list.

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