Welcome to Subscribe On Youtube

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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
          ]
  
  

• 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)
  }
  

• 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);
  }
  
  

• 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
      }
  }
  
  

All Problems

All Solutions