What is the time complexity of Count Integers With Even Digit Sum?

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

What data structures are used to solve Count Integers With Even Digit Sum?

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

What are the key insights for solving Count Integers With Even Digit Sum?

Insight 1: We need to iterate through each number from 1 to `num` and calculate the sum of its digits. Insight 2: A simple iterative approach is sufficient due to the constraint that `num` is at most 1000.

Which companies ask LeetCode 2180. Count Integers With Even Digit Sum in interviews?

This problem is frequently asked at MindTree.

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Count Integers With Even Digit Sum - LeetCode 2180 Solution

Count Integers With Even Digit Sum - Complete Solution Guide

Count Integers With Even Digit Sum is LeetCode problem 2180, 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 a positive integer num , return the number of positive integers less than or equal to num whose digit sums are even . The digit sum of a positive integer is the sum of all its digits. Example 1: Input: num = 4 Output: 2 Explanation: The only integers less than or equal to 4 whose digit sums are even are 2 and 4. Example 2: Input: num = 30 Output: 14 Explanation: The 14 integers less than or equal to 30 whose digit sums are even are 2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28. Co

Detailed Explanation

The problem asks you to count the number of positive integers from 1 up to a given integer `num` (inclusive) where the sum of their digits is even. For example, if `num` is 4, the integers are 1, 2, 3, 4. The digit sums are 1, 2, 3, 4 respectively. Only 2 and 4 have even digit sums, so the answer is 2. The input `num` is guaranteed to be between 1 and 1000.

Solution Approach

The provided solutions use a brute-force approach. They iterate through each number from 1 to `num`. For each number, they calculate the sum of its digits. If this sum is even, a counter is incremented. Finally, the counter (representing the number of integers with even digit sums) is returned.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0.
Step 2: Iterate through numbers from 1 to `num` (inclusive).
Step 3: For each number, calculate the sum of its digits. This can be done by converting the number to a string, iterating through its digits, converting them back to integers and summing them, or by repeatedly using the modulo operator (%) and integer division (/).
Step 4: Check if the sum of digits is even using the modulo operator (`%`).
Step 5: If the sum is even, increment the `count`.
Step 6: After iterating through all numbers, return the final value of `count`.

Key Insights

Insight 1: We need to iterate through each number from 1 to `num` and calculate the sum of its digits.
Insight 2: A simple iterative approach is sufficient due to the constraint that `num` is at most 1000.
Insight 3: No significant optimizations are needed for this problem given the constraints, though a mathematical approach could improve performance for much larger inputs.

Complexity Analysis

Time Complexity: O(n*log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Simulation.

Companies

Asked at: MindTree.

Count Integers With Even Digit Sum | LeetCode 2180

Easy

Math Simulation

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

Solution Code

Available in:

class Solution {
    public int countEven(int num) {
        int count = 0;
        for (int i = 1; i <= num; i++) {
            int sum = 0;
            int temp = i;
            while (temp > 0) {
                sum += temp % 10;
                temp /= 10;
            }
            if (sum % 2 == 0) {
                count++;
            }
        }
        return count;
    }
}

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Problem Info

Description

Problem 2180: Count Integers With Even Digit Sum - Detailed explanation and solution approaches.

Topics

Math Simulation

Companies

MindTree

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2180

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

6 steps

1
Step 1: Initialize a counter `count` to 0.
2
Step 2: Iterate through numbers from 1 to `num` (inclusive).
3
Step 3: For each number, calculate the sum of its digits. This can be done by converting the number to a string, iterating through its digits, converting them back to integers and summing them, or by repeatedly using the modulo operator (%) and integer division (/).
4
Step 4: Check if the sum of digits is even using the modulo operator (`%`).
5
Step 5: If the sum is even, increment the `count`.
6
Step 6: After iterating through all numbers, return the final 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(n*log(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n * log(n)) because we iterate `n` times (where n = num), and the digit sum calculation for each number takes time proportional to the number of digits (which is at most log10(n)). The space complexity is O(1) because we only use a constant amount of extra space to store variables like `count`, `sum`, and `temp`.

📖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

Mathematical Approach

For much larger values of `num`, a mathematical approach could be more efficient. We could potentially find a pattern in the number of integers with even digit sums within ranges of powers of 10 and use that to calculate the result faster without iterating through each number.

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

Trade-offs: The mathematical approach would be significantly faster for very large values of `num`, but it would require more complex calculations to establish the pattern. For the given constraint (num <= 1000), the simple iterative approach is perfectly acceptable.

✨

Key Insights

1Insight 1: We need to iterate through each number from 1 to `num` and calculate the sum of its digits.
2Insight 2: A simple iterative approach is sufficient due to the constraint that `num` is at most 1000.
3Insight 3: No significant optimizations are needed for this problem given the constraints, though a mathematical approach could improve performance for much larger inputs.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `num` is 1. The solution correctly handles this, returning 0 because the digit sum of 1 is odd.
Edge case 2: `num` is a power of 10 (e.g., 10, 100, 1000). The solution correctly handles these cases by iterating through all numbers and checking their digit sums.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the sum of digits. Beginners might make mistakes with the modulo operator or integer division.

✗Common mistake 2: Off-by-one errors in the loop boundaries. Forgetting to include or exclude `num` in the iteration.

✅

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: This type of digit sum calculation could be used in cryptography or other number theory related applications.

🏦Real-world scenario 2: This problem highlights the importance of understanding how to analyze and solve problems using simple iterative techniques, which are commonly used in various programming scenarios.

💡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.

👨‍💻 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 21, 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 14

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

@AlgoMasterJul 7

❓

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 Integers With Even Digit Sum"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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