How do you solve Simplified Fractions on LeetCode?

Simplified Fractions (LeetCode #1447) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding GCD is crucial for identifying simplified fractions.

What is the brute force solution for Simplified Fractions?

The brute force approach for Simplified Fractions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates all possible fractions with denominators from 2 to n and checks if they are simplified. This method is straightforward but inefficient for larger values of n. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Simplified Fractions?

Simplified Fractions uses the following concepts: Math, String, Number Theory. The recommended patterns to study are: Number Theory, GCD, Fractions.

What are the interview tips for Simplified Fractions?

Always clarify the constraints and edge cases with the interviewer. Explain your thought process as you write code to demonstrate understanding. Consider both brute force and optimized approaches to show depth of knowledge.

What are common mistakes when solving Simplified Fractions?

Not checking for GCD correctly, leading to incorrect fractions. Overlooking the range of numerators and denominators.

#1447

Simplified Fractions

Medium

MathStringNumber TheoryNumber TheoryGCDFractions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

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Intuition

Time O(n²)Space O(n)

The optimal solution leverages the properties of fractions and uses a mathematical approach to generate only the simplified fractions without checking each one. This is more efficient and avoids unnecessary calculations.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store the simplified fractions.
2Step 2: Loop through all denominators from 2 to n.
3Step 3: For each denominator, loop through numerators from 1 to (denominator - 1).
4Step 4: Use the GCD to check if the fraction is simplified. If GCD(numerator, denominator) is 1, add it to the list.
5Step 5: Return the list of simplified fractions.

solution.py5 lines

1# Full working Python code
2from math import gcd
3
4def simplifiedFractions(n):
5    return [f'{numerator}/{denominator}' for denominator in range(2, n + 1) for numerator in range(1, denominator) if gcd(numerator, denominator) == 1]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we are more efficient in checking and storing results. The space complexity is O(n) as we store the results in a list.

1Understanding GCD is crucial for identifying simplified fractions.
2Recognizing patterns in fraction generation helps optimize the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.