Count Integers With Even Digit Sum

Count Integers With Even Digit Sum - Problem

Math Simulation Easy

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.

Input & Output

Example 1 — Small Number

$ Input: num = 4

› Output: 2

💡 Note: Check numbers 1-4: 1(sum=1,odd), 2(sum=2,even), 3(sum=3,odd), 4(sum=4,even). Count of even sums = 2.

Example 2 — Larger Number

$ Input: num = 30

› Output: 14

💡 Note: Among numbers 1-30, exactly 14 have even digit sums (like 2,4,6,8,11,13,15,17,20,22,24,26,28,29).

Example 3 — Single Digit

$ Input: num = 9

› Output: 4

💡 Note: Numbers 1-9 with even digit sums: 2,4,6,8. Total count = 4.

Constraints

1 ≤ num ≤ 1000

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is recognizing that digit sum parity follows mathematical patterns. The optimal approach iterates through all numbers calculating digit sums. Time: O(n × log n), Space: O(1).

Common Approaches

✓ Brute Force - Check Every Number

⏱️ Time: O(n × log n) Space: O(1)

For each number from 1 to num, calculate the sum of its digits by repeatedly taking modulo 10 and dividing by 10. Count how many have even digit sums.

Mathematical Pattern Recognition

⏱️ Time: O(log n) Space: O(1)

By mathematical analysis, in any sufficiently large range, roughly half the numbers will have even digit sums and half will have odd digit sums due to the balanced distribution of digits.

Brute Force - Check Every Number — Algorithm Steps

Iterate from 1 to num
For each number, calculate digit sum by extracting digits
Count numbers with even digit sums

Visualization

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Step-by-Step Walkthrough

Check Each Number

Go through numbers 1, 2, 3, 4...

Calculate Digit Sum

Add up all digits: 12 → 1+2=3

Count Even Sums

Increment counter if sum is even

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int digitSum(int n) {
    int total = 0;
    while (n > 0) {
        total += n % 10;
        n /= 10;
    }
    return total;
}

int solution(int num) {
    int count = 0;
    for (int i = 1; i <= num; i++) {
        if (digitSum(i) % 2 == 0) {
            count++;
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n × log n)

Iterate through n numbers, each taking O(log n) time to calculate digit sum

⚡ Linearithmic

Space Complexity

O(1)

Only using variables to track count and current number

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Every Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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