How do you solve Fraction Addition and Subtraction on LeetCode?

Fraction Addition and Subtraction (LeetCode #592) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding how to handle fractions is crucial in many mathematical problems.

What is the time complexity of Fraction Addition and Subtraction?

The optimal solution for Fraction Addition and Subtraction runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the string once to extract fractions and compute the result, making it efficient.

What is the brute force solution for Fraction Addition and Subtraction?

The brute force approach for Fraction Addition and Subtraction is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves parsing the string expression to extract each fraction, converting them to a common denominator, and then performing the addition or subtraction. This method is straightforward but inefficient for larger inputs due to repeated calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fraction Addition and Subtraction?

Fraction Addition and Subtraction uses the following concepts: Math, String, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Fraction Addition and Subtraction?

Always clarify the input format and constraints before starting. Think about edge cases, like zero results or large denominators. Practice breaking down the problem into smaller parts.

What are common mistakes when solving Fraction Addition and Subtraction?

Forgetting to handle signs correctly when parsing fractions. Not simplifying the final result.

#592

Fraction Addition and Subtraction

Medium

MathStringSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution efficiently processes the expression in a single pass, directly calculating the total numerator and denominator without needing to repeatedly find a common denominator. This reduces unnecessary computations and speeds up the process.

⚙️

Algorithm

5 steps

1Step 1: Initialize total numerator and denominator to 0 and 1 respectively.
2Step 2: Split the expression into fractions using regex to capture signs and numerators/denominators.
3Step 3: For each fraction, update the total numerator and denominator using the formula for adding fractions.
4Step 4: After processing all fractions, simplify the result using the GCD.
5Step 5: Format the result as a string in the form 'numerator/denominator'.

solution.py20 lines

1from math import gcd
2
3def fractionAddition(expression):
4    total_num = 0
5    total_denom = 1
6    fractions = expression.split(' ')
7    for frac in fractions:
8        sign = 1
9        if frac[0] == '-':
10            sign = -1
11            frac = frac[1:]
12        elif frac[0] == '+':
13            frac = frac[1:]
14        num, denom = map(int, frac.split('/'))
15        total_num = total_num * denom + sign * num * total_denom
16        total_denom *= denom
17    if total_num == 0:
18        return '0/1'
19    divisor = gcd(total_num, total_denom)
20    return f'{total_num // divisor}/{total_denom // divisor}'

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once to extract fractions and compute the result, making it efficient.

1Understanding how to handle fractions is crucial in many mathematical problems.
2Simplifying fractions using GCD is a common technique.

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