Count Integers With Even Digit Sum - Practice Coding Problems

Count Integers With Even Digit Sum - Problem

Math Simulation Easy

Given a positive integer num, you need to count how many positive integers from 1 to num (inclusive) have an even digit sum.

The digit sum of a positive integer is calculated by adding all its individual digits together. For example:

123 → digit sum = 1 + 2 + 3 = 6 (even)
45 → digit sum = 4 + 5 = 9 (odd)
8 → digit sum = 8 (even)

Goal: Return the count of all positive integers ≤ num whose digit sums are even.

Input: A positive integer num
Output: An integer representing the count of numbers with even digit sums

Input & Output

example_1.py — Basic Case

$ Input: num = 4

› Output: 2

💡 Note: Numbers 1 to 4: 1(sum=1,odd), 2(sum=2,even), 3(sum=3,odd), 4(sum=4,even). Count = 2

example_2.py — Larger Number

$ Input: num = 30

› Output: 14

💡 Note: From 1 to 30, the numbers with even digit sums are: 2,4,6,8,11,13,15,17,19,20,22,24,26,28. Total count = 14

example_3.py — Single Digit

$ Input: num = 1

› Output: 0

💡 Note: Only number 1 exists, its digit sum is 1 (odd), so count = 0

Constraints

1 ≤ num ≤ 1000
num is a positive integer
Digit sum is calculated by adding all digits of a number

Visualization

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Understanding the Visualization

Initialize Counter

Start with count = 0 to track numbers with even digit sums

Process Each Number

For each number from 1 to num, extract and sum all digits

Check Even/Odd

If digit sum is divisible by 2, increment our counter

Return Result

After checking all numbers, return the final count

Key Takeaway

🎯 Key Insight: This is a simulation problem that requires checking each number individually - there's no mathematical shortcut to avoid calculating digit sums

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 Apple 6

This problem requires iterating through numbers 1 to num and calculating digit sums. The optimal approach uses simulation with O(n * log n) time complexity, where we check each number and sum its digits to determine if the sum is even.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(n * log(n))	O(1)	Check each number from 1 to num and calculate digit sum

Brute Force Simulation — Algorithm Steps

Initialize counter to 0
Loop through each number i from 1 to num
For each number, extract digits and sum them
If digit sum is even, increment counter
Return the final counter

Visualization

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Step-by-Step Walkthrough

Start with num=4

We need to check numbers 1, 2, 3, 4

Check number 1

Digit sum = 1 (odd), don't count

Check number 2

Digit sum = 2 (even), count = 1

Check number 3

Digit sum = 3 (odd), don't count

Check number 4

Digit sum = 4 (even), count = 2

Code -

solution.c — C

#include <stdio.h>

int countEvenDigitSum(int num) {
    int count = 0;
    for (int i = 1; i <= num; i++) {
        int digitSum = 0;
        int temp = i;
        while (temp > 0) {
            digitSum += temp % 10;
            temp /= 10;
        }
        if (digitSum % 2 == 0) {
            count++;
        }
    }
    return count;
}

int main() {
    int num;
    scanf("%d", &num);
    printf("%d\n", countEvenDigitSum(num));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n * log(n))

We iterate through n numbers, and for each number, we need O(log(n)) time to extract and sum its digits

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for counting and temporary calculations

✓ Linear Space

28.6K Views

Medium Frequency

~8 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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