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1447. Simplified Fractions

Description

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. You can return the answer in any order.

 

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

Solutions

Solution 1

Thinking

\(n\le 100\). Enumerate \(1\le i<j\le n\) and keep the pair when \(\gcd(i,j)=1\).

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class Solution:
    def simplifiedFractions(self, n: int) -> List[str]:
        return [
            f'{i}/{j}'
            for i in range(1, n)
            for j in range(i + 1, n + 1)
            if gcd(i, j) == 1
        ]

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class Solution {
    public List<String> simplifiedFractions(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i < n; ++i) {
            for (int j = i + 1; j < n + 1; ++j) {
                if (gcd(i, j) == 1) {
                    ans.add(i + "/" + j);
                }
            }
        }
        return ans;
    }

    private int gcd(int a, int b) {
        return b > 0 ? gcd(b, a % b) : a;
    }
}

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class Solution {
public:
    vector<string> simplifiedFractions(int n) {
        vector<string> ans;
        for (int i = 1; i < n; ++i) {
            for (int j = i + 1; j < n + 1; ++j) {
                if (__gcd(i, j) == 1) {
                    ans.push_back(to_string(i) + "/" + to_string(j));
                }
            }
        }
        return ans;
    }
};

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func simplifiedFractions(n int) (ans []string) {
    for i := 1; i < n; i++ {
        for j := i + 1; j < n+1; j++ {
            if gcd(i, j) == 1 {
                ans = append(ans, strconv.Itoa(i)+"/"+strconv.Itoa(j))
            }
        }
    }
    return ans
}

func gcd(a, b int) int {
    if b == 0 {
        return a
    }
    return gcd(b, a%b)
}

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function simplifiedFractions(n: number): string[] {
    const ans: string[] = [];
    for (let i = 1; i < n; ++i) {
        for (let j = i + 1; j < n + 1; ++j) {
            if (gcd(i, j) === 1) {
                ans.push(`${i}/${j}`);
            }
        }
    }
    return ans;
}

function gcd(a: number, b: number): number {
    return b === 0 ? a : gcd(b, a % b);
}

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impl Solution {
    fn gcd(a: i32, b: i32) -> i32 {
        match b {
            0 => a,
            _ => Solution::gcd(b, a % b),
        }
    }

    pub fn simplified_fractions(n: i32) -> Vec<String> {
        let mut res = vec![];
        for i in 1..n {
            for j in i + 1..=n {
                if Solution::gcd(i, j) == 1 {
                    res.push(format!("{}/{}", i, j));
                }
            }
        }
        res
    }
}

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