What is the time complexity of Fraction to Recurring Decimal?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Fraction to Recurring Decimal?

This problem uses Hash Table, Math, String. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Fraction to Recurring Decimal?

Insight 1: The core idea is to simulate long division to obtain the decimal representation. Insight 2: A hash map (or array in C) is crucial to detect repeating remainders. If a remainder repeats, the decimal digits generated from that point onwards also repeat.

Which companies ask LeetCode 166. Fraction to Recurring Decimal in interviews?

This problem is frequently asked at Airbnb, Goldman Sachs, Google, IXL, ServiceNow and 1 more companies.

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Fraction to Recurring Decimal - LeetCode 166 Solution

Fraction to Recurring Decimal - Complete Solution Guide

Fraction to Recurring Decimal is LeetCode problem 166, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format . If the fractional part is repeating, enclose the repeating part in parentheses If multiple answers are possible, return any of them . It is guaranteed that the length of the answer string is less than 10 4 for all the given inputs. Note that if the fraction can be represented as a finite length string , you must return it. Example 1: Input: numerator = 1, denominator = 2 Output: "0

Detailed Explanation

The problem asks us to convert a fraction, represented by its numerator and denominator, into a string format. If the decimal representation has a repeating part, we need to enclose that repeating part in parentheses. For instance, 1/2 becomes "0.5", 2/1 becomes "2", and 4/333 becomes "0.(012)". We are guaranteed that the length of the resulting string is less than 10000. It's important to handle potential negative inputs for both the numerator and denominator. The problem requires handling both integer results (no decimal part) and decimal results that either terminate or recur.

Solution Approach

The provided solutions simulate long division. First, determine the sign of the result. Then calculate the integer part of the fraction. If the remainder is 0 at this point, the fraction is a whole number, and we are done. Otherwise, start calculating the decimal part. Use a hash map to store each remainder encountered and its corresponding index in the result string. If a remainder repeats, it means the decimal part from that index repeats, and we enclose it in parentheses.

Step-by-Step Algorithm

Step 1: Handle the sign. Determine if the result should be negative by checking if the numerator and denominator have different signs.
Step 2: Convert numerator and denominator to their absolute values (long type) to avoid potential integer overflow during calculations.
Step 3: Calculate the integer part of the fraction (numerator // denominator) and append it to the result string.
Step 4: Calculate the initial remainder (numerator % denominator). If the remainder is 0, the fraction is an integer, so return the result.
Step 5: If the remainder is not 0, append a decimal point "." to the result string.
Step 6: Initialize a hash map (or array in C) to store remainders and their corresponding indices in the result string.
Step 7: Enter a loop that continues as long as the remainder is not 0.
Step 8: Inside the loop, check if the current remainder is already in the hash map. If it is, it means we have found a repeating decimal. Insert a "(" at the index stored in the hash map and append a ")" at the end of the result string, then break out of the loop.
Step 9: If the remainder is not in the hash map, store the remainder and its current index in the result string into the hash map.
Step 10: Multiply the remainder by 10 to get the next digit in the decimal part.
Step 11: Calculate the next digit (remainder // denominator) and append it to the result string.
Step 12: Update the remainder (remainder % denominator) for the next iteration.
Step 13: After the loop finishes (either by finding a repeating decimal or the remainder becoming 0), return the result string.

Key Insights

Insight 1: The core idea is to simulate long division to obtain the decimal representation.
Insight 2: A hash map (or array in C) is crucial to detect repeating remainders. If a remainder repeats, the decimal digits generated from that point onwards also repeat.
Insight 3: We need to handle negative numerator or denominator. To avoid integer overflow issues, especially when numerator and denominator are close to the limits of integer range, converting them to long type before taking absolute value is important.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Math, String.

Companies

Asked at: Airbnb, Goldman Sachs, Google, IXL, ServiceNow, TikTok.

Fraction to Recurring Decimal | LeetCode 166

Medium

Hash Table Math String

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) {
            return "0";
        }

        StringBuilder result = new StringBuilder();
        if ((numerator < 0) != (denominator < 0)) {
            result.append("-");
        }

        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);

        // Integer part
        result.append(String.valueOf(num / den));
        long remainder = num % den;

        if (remainder == 0) {
            return result.toString();
        }

        // Fractional part
        result.append(".");
        Map<Long, Integer> remainderMap = new HashMap<>();

        while (remainder != 0) {
            if (remainderMap.containsKey(remainder)) {
                int startIndex = remainderMap.get(remainder);
                result.insert(startIndex, "(");
                result.append(")");
                break;
            }
            
            remainderMap.put(remainder, result.length());
            
            remainder *= 10;
            result.append(String.valueOf(remainder / den));
            remainder %= den;
        }
        
        return result.toString();
    }
}

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Problem Info

Description

Problem 166: Fraction to Recurring Decimal - Detailed explanation and solution approaches.

Topics

Hash Table Math String

Companies

AirbnbGoldman SachsGoogleIXLServiceNowTikTok

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:166

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Handle the sign. Determine if the result should be negative by checking if the numerator and denominator have different signs.
2
Step 2: Convert numerator and denominator to their absolute values (long type) to avoid potential integer overflow during calculations.
3
Step 3: Calculate the integer part of the fraction (numerator // denominator) and append it to the result string.
4
Step 4: Calculate the initial remainder (numerator % denominator). If the remainder is 0, the fraction is an integer, so return the result.
5
Step 5: If the remainder is not 0, append a decimal point "." to the result string.
6
Step 6: Initialize a hash map (or array in C) to store remainders and their corresponding indices in the result string.
7
Step 7: Enter a loop that continues as long as the remainder is not 0.
8
Step 8: Inside the loop, check if the current remainder is already in the hash map. If it is, it means we have found a repeating decimal. Insert a "(" at the index stored in the hash map and append a ")" at the end of the result string, then break out of the loop.
9
Step 9: If the remainder is not in the hash map, store the remainder and its current index in the result string into the hash map.
10
Step 10: Multiply the remainder by 10 to get the next digit in the decimal part.
11
Step 11: Calculate the next digit (remainder // denominator) and append it to the result string.
12
Step 12: Update the remainder (remainder % denominator) for the next iteration.
13
Step 13: After the loop finishes (either by finding a repeating decimal or the remainder becoming 0), return the result string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the while loop. The number of iterations is proportional to the length of the decimal representation. The space complexity is determined by the hash map which stores remainders. In the worst case scenario, we need to store all the remainders encountered before encountering a repeating pattern, resulting in O(n) space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using BigInteger/BigDecimal (Java)

For languages like Java, you could potentially use BigInteger or BigDecimal to handle arbitrarily large numerators and denominators without worrying about integer overflows. However, simulating the long division process is more efficient in this case as it allows the detection of repeating patterns earlier.

Time: Potentially higher than O(n) due to the general overhead of BigInteger/BigDecimal operationsSpace: Potentially higher than O(n) depending on the implementation details

Trade-offs: This approach could be simpler in terms of code readability but might be less efficient. Useful if the inputs are truly unbounded and avoiding overflow is the primary concern, not performance.

✨

Key Insights

1Insight 1: The core idea is to simulate long division to obtain the decimal representation.
2Insight 2: A hash map (or array in C) is crucial to detect repeating remainders. If a remainder repeats, the decimal digits generated from that point onwards also repeat.
3Insight 3: We need to handle negative numerator or denominator. To avoid integer overflow issues, especially when numerator and denominator are close to the limits of integer range, converting them to long type before taking absolute value is important.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: numerator = 0. The result is simply "0". All solutions correctly handle this.
Edge case 2: denominator = 1. The result is simply the numerator. All solutions correctly handle this.
Edge case 3: Finite decimal representation (e.g., 1/2). The algorithm correctly terminates and returns the finite decimal.
Edge case 4: Integer overflow. When numerator or denominator are Integer.MIN_VALUE. Converting the numerator and denominator to `long` type before taking absolute value avoids integer overflow. All solutions correctly address this.
Edge case 5: Negative numerator or denominator. The algorithm correctly handles this by determining the sign of the result at the beginning.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the sign of the resulting fraction.

✗Common mistake 2: Integer overflow issues when taking the absolute value of Integer.MIN_VALUE. Convert to `long` before taking the absolute value.

✗Common mistake 3: Not detecting the repeating decimal part correctly, or getting the starting index wrong.

✗Common mistake 4: Failing to handle the case when the remainder is 0, resulting in an infinite loop or incorrect output.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial calculations where fractions need to be represented accurately, and recurring decimals must be identified to avoid rounding errors.

🏦Real-world scenario 2: Scientific computing where precise fractional representation is crucial for simulations and data analysis. This can be used in unit conversion or any kind of calculation that might result in a repeating decimal.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Fraction to Recurring Decimal"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?