What is the time complexity of Self Dividing Numbers?

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

What data structures are used to solve Self Dividing Numbers?

This problem uses Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Self Dividing Numbers?

Insight 1: The core idea is to iterate through the numbers in the given range and check each number for the self-dividing property. Insight 2: To check if a number is self-dividing, we need to extract its digits and check if each digit divides the original number without any remainder and that each digit is non-zero.

Which companies ask LeetCode 728. Self Dividing Numbers in interviews?

This problem is frequently asked at Epic Systems.

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Self Dividing Numbers - LeetCode 728 Solution

Self Dividing Numbers - Complete Solution Guide

Self Dividing Numbers is LeetCode problem 728, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A self-dividing number is a number that is divisible by every digit it contains. For example, 128 is a self-dividing number because 128 % 1 == 0 , 128 % 2 == 0 , and 128 % 8 == 0 . A self-dividing number is not allowed to contain the digit zero. Given two integers left and right , return a list of all the self-dividing numbers in the range [left, right] (both inclusive ). Example 1: Input: left = 1, right = 22 Output: [1,2,3,4,5,6,7,8,9,11,12,15,22] Example 2: Input: left = 47, right = 85 Output

Detailed Explanation

The problem asks us to find all self-dividing numbers within a given range [left, right] (inclusive). A self-dividing number is defined as a number that is divisible by every digit it contains. Additionally, a self-dividing number cannot contain the digit zero.

Solution Approach

The solution involves iterating through the numbers from 'left' to 'right'. For each number, we check if it's a self-dividing number. The checking is done by extracting each digit from the number. If a digit is zero or if the number is not divisible by the digit, the number is not self-dividing. If all digits pass the divisibility test, the number is added to the result list.

Step-by-Step Algorithm

Step 1: Initialize an empty list (result) to store the self-dividing numbers.
Step 2: Iterate through the numbers from 'left' to 'right' (inclusive).
Step 3: For each number 'i', check if it's self-dividing.
Step 4: To check if 'i' is self-dividing, extract each of its digits.
Step 5: For each extracted digit, check if the digit is zero or if 'i' is not divisible by the digit. If either condition is true, 'i' is not self-dividing, and the inner loop is broken.
Step 6: If all digits of 'i' divide 'i' without remainder and are not zero, 'i' is self-dividing. Append 'i' to the 'result' list.
Step 7: After iterating through all numbers in the range, return the 'result' list.

Key Insights

Insight 1: The core idea is to iterate through the numbers in the given range and check each number for the self-dividing property.
Insight 2: To check if a number is self-dividing, we need to extract its digits and check if each digit divides the original number without any remainder and that each digit is non-zero.
Insight 3: Converting the integer to a string simplifies the digit extraction process in some languages like Python. In other languages like Java/C/C++, we use modulo and division operations to extract digits.

Complexity Analysis

Time Complexity: O(n*k)

Space Complexity: O(n)

Topics

This problem involves: Math.

Companies

Asked at: Epic Systems.

Self Dividing Numbers | LeetCode 728

Easy

Math

Time: O(n*k)Space: O(n)

Solution Code

Available in:

import java.util.*;
class Solution {
    public List<Integer> selfDividingNumbers(int left, int right) {
        List<Integer> result = new ArrayList<>();
        for (int i = left; i <= right; i++) {
            if (isSelfDividing(i)) {
                result.add(i);
            }
        }
        return result;
    }

    private boolean isSelfDividing(int n) {
        int temp = n;
        while (temp > 0) {
            int digit = temp % 10;
            if (digit == 0 || n % digit != 0) {
                return false;
            }
            temp /= 10;
        }
        return true;
    }
}

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Problem Info

Description

Problem 728: Self Dividing Numbers - Detailed explanation and solution approaches.

Topics

Math

Companies

Epic Systems

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:728

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

7 steps

1
Step 1: Initialize an empty list (result) to store the self-dividing numbers.
2
Step 2: Iterate through the numbers from 'left' to 'right' (inclusive).
3
Step 3: For each number 'i', check if it's self-dividing.
4
Step 4: To check if 'i' is self-dividing, extract each of its digits.
5
Step 5: For each extracted digit, check if the digit is zero or if 'i' is not divisible by the digit. If either condition is true, 'i' is not self-dividing, and the inner loop is broken.
6
Step 6: If all digits of 'i' divide 'i' without remainder and are not zero, 'i' is self-dividing. Append 'i' to the 'result' list.
7
Step 7: After iterating through all numbers in the range, return the 'result' list.

💡

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*k)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The outer loop iterates 'n' times, where 'n' is the size of the range. The inner loop (digit extraction and divisibility check) iterates 'k' times, where 'k' is the number of digits in a number. The space complexity is O(n) because we create a list to store the results and, in the worst case, the size of the result list could be equal to the size of the range.

📖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

Precomputation

Precompute self-dividing numbers up to 10000 and store them in a set. Then, for the given range [left, right], simply filter the precomputed set for numbers within that range.

Time: O(1) for the given constraints. O(N) for the initial precomputation, where N=10000.Space: O(M) where M is the count of self-dividing numbers until 10000, which would be a constant.

Trade-offs: This approach is more efficient if the function is called multiple times with different ranges within the constraint 1 <= left <= right <= 10000, as it avoids repeated calculations. The initial precomputation cost is amortized over multiple calls.

✨

Key Insights

1Insight 1: The core idea is to iterate through the numbers in the given range and check each number for the self-dividing property.
2Insight 2: To check if a number is self-dividing, we need to extract its digits and check if each digit divides the original number without any remainder and that each digit is non-zero.
3Insight 3: Converting the integer to a string simplifies the digit extraction process in some languages like Python. In other languages like Java/C/C++, we use modulo and division operations to extract digits.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The input range is very small, like [1, 9]. The algorithm correctly identifies all single-digit numbers as self-dividing.
Edge case 2: The input range includes numbers with zero as a digit. The algorithm correctly identifies these numbers as not self-dividing.
Edge case 3: The input range has identical left and right values e.g. [10, 10]. The algorithm correctly checks and determines if the number is self-dividing, and returns the list [10] if it is, and [] otherwise.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a digit is zero. This would lead to a division by zero error or incorrect results.

✗Common mistake 2: Incorrectly extracting the digits from the number. Using string conversion is generally safer, but integer division and modulo operations are also viable.

✗Common mistake 3: Not including the 'right' endpoint in the iteration range, resulting in missing self-dividing numbers at the upper bound.

✅

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: Data validation. Self-dividing number logic can be applied to validate data entries to ensure that values meet certain divisibility criteria.

🏦Real-world scenario 2: Number theory research. Identifying and classifying self-dividing numbers contributes to research in number theory and pattern recognition in numerical sequences.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary 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: June 9, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Self Dividing Numbers"?

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

🤖 Need help with this problem?