What is the time complexity of Count the Digits That Divide a Number?

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

What data structures are used to solve Count the Digits That Divide a Number?

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

What are the key insights for solving Count the Digits That Divide a Number?

Insight 1: The problem requires iterating through each digit of the input number. This can be efficiently done using modulo (%) and integer division (//) operations. Insight 2: Extracting digits can be done by repeatedly taking the remainder when dividing by 10, then integer dividing by 10 to remove the last digit.

Which companies ask LeetCode 2520. Count the Digits That Divide a Number in interviews?

This problem is frequently asked at tcs.

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Count the Digits That Divide a Number - LeetCode 2520 Solution

Count the Digits That Divide a Number - Complete Solution Guide

Count the Digits That Divide a Number is LeetCode problem 2520, 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

Given an integer num , return the number of digits in num that divide num . An integer val divides nums if nums % val == 0 . Example 1: Input: num = 7 Output: 1 Explanation: 7 divides itself, hence the answer is 1. Example 2: Input: num = 121 Output: 2 Explanation: 121 is divisible by 1, but not 2. Since 1 occurs twice as a digit, we return 2. Example 3: Input: num = 1248 Output: 4 Explanation: 1248 is divisible by all of its digits, hence the answer is 4. Constraints: 1 <= num <= 10 9 num does

Detailed Explanation

The problem asks you to count how many digits of a given integer `num` divide the integer evenly. A digit divides the integer if the remainder of the integer divided by that digit is 0. The input is a positive integer `num` (between 1 and 10<sup>9</sup> inclusive) that does not contain the digit 0. The output is the count of digits in `num` that divide `num` without leaving a remainder.

Solution Approach

The provided code uses a iterative approach. It extracts each digit from the input number `num` one by one. For each digit, it checks if `num` is divisible by that digit. If it is, a counter is incremented. This process continues until all digits have been checked. Finally, the counter, representing the number of digits that divide `num`, is returned.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0. This variable will store the number of digits that divide `num`.
Step 2: Create a temporary variable `temp` and assign it the value of `num`. This is done to avoid modifying the original `num` during the digit extraction process.
Step 3: Enter a `while` loop that continues as long as `temp` is greater than 0. This loop iterates through each digit of the number.
Step 4: Extract the last digit of `temp` using the modulo operator (`temp % 10`) and store it in the `digit` variable.
Step 5: Check if `num` is divisible by `digit` using the modulo operator (`num % digit == 0`). If it is, increment the `count`.
Step 6: Remove the last digit from `temp` using integer division (`temp //= 10`).
Step 7: Repeat steps 4-6 until `temp` becomes 0.
Step 8: After the loop finishes, return the value of `count`.

Key Insights

Insight 1: The problem requires iterating through each digit of the input number. This can be efficiently done using modulo (%) and integer division (//) operations.
Insight 2: Extracting digits can be done by repeatedly taking the remainder when dividing by 10, then integer dividing by 10 to remove the last digit.
Insight 3: The constraint that `num` does not contain 0 simplifies the code, as we don't need to handle division by zero errors.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: tcs.

Count the Digits That Divide a Number | LeetCode 2520

Easy

Math

Time: O(log(n))Space: O(1)

Solution Code

Available in:

class Solution {
    public int countDigits(int num) {
        int count = 0;
        int temp = num;
        while (temp > 0) {
            int digit = temp % 10;
            if (num % digit == 0) {
                count++;
            }
            temp /= 10;
        }
        return count;
    }
}

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Problem Info

Description

Problem 2520: Count the Digits That Divide a Number - Detailed explanation and solution approaches.

Topics

Math

Companies

tcs

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2520

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

8 steps

1
Step 1: Initialize a counter `count` to 0. This variable will store the number of digits that divide `num`.
2
Step 2: Create a temporary variable `temp` and assign it the value of `num`. This is done to avoid modifying the original `num` during the digit extraction process.
3
Step 3: Enter a `while` loop that continues as long as `temp` is greater than 0. This loop iterates through each digit of the number.
4
Step 4: Extract the last digit of `temp` using the modulo operator (`temp % 10`) and store it in the `digit` variable.
5
Step 5: Check if `num` is divisible by `digit` using the modulo operator (`num % digit == 0`). If it is, increment the `count`.
6
Step 6: Remove the last digit from `temp` using integer division (`temp //= 10`).
7
Step 7: Repeat steps 4-6 until `temp` becomes 0.
8
Step 8: After the loop finishes, return the value of `count`.

💡

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(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because the number of iterations in the `while` loop is proportional to the number of digits in `num`. The number of digits in a number `num` is approximately log<sub>10</sub>(num). The space complexity is constant because we only use a few integer variables to store the input number, temporary values, and the count. The space used does not depend on the size of the input number.

📖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

✨

Key Insights

1Insight 1: The problem requires iterating through each digit of the input number. This can be efficiently done using modulo (%) and integer division (//) operations.
2Insight 2: Extracting digits can be done by repeatedly taking the remainder when dividing by 10, then integer dividing by 10 to remove the last digit.
3Insight 3: The constraint that `num` does not contain 0 simplifies the code, as we don't need to handle division by zero errors.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `num` is a single-digit number. The solution correctly handles this by performing the division and incrementing the count only once.
Edge case 2: `num` is a number where all digits divide `num` (e.g., 1248). The solution correctly counts all such digits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to create a temporary variable to store a copy of `num`. Modifying `num` directly within the loop will lead to incorrect results.

✗Common mistake 2: Incorrectly using the modulo operator or integer division, leading to incorrect digit extraction or divisibility checks.

✅

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: Analyzing numerical patterns in datasets to find specific divisibility properties. This could be used in cryptography or number theory research.

🏦Real-world scenario 2: In data validation, checking if numbers meet specific divisibility rules (e.g., check digit algorithms in credit card numbers).

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

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 1, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

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

@AlgoMasterJul 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 "Count the Digits That Divide a Number"?

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.

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