Count Integers With Even Digit Sum

LeetCode 2298 | Difficulty: Easy

Easy

Problem Description

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.

Constraints:

- `1 <= num <= 1000`

Topics: Math, Simulation

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Approach

Mathematical

Look for mathematical patterns or formulas. Consider: modular arithmetic, GCD/LCM, prime factorization, combinatorics, or geometric properties.

When to use

Problems with clear mathematical structure, counting, number properties.

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Solutions

Solution 1: C# (Best: 21 ms)

Metric	Value
Runtime	21 ms
Memory	25.1 MB
Date	2022-02-21

Solution

public class Solution {
    public int CountEven(int num) {
        int i = 2, result = 0;
        while(i<=num)
        {
            int temp = i; int sum = 0;
            while(temp>0)
            {
                sum += temp%10;
                temp/=10;
            }
            if(sum%2 ==0) result++;
            i++;
        }
        
        return  result;
    }
}

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Complexity Analysis

Approach	Time	Space
Solution	$O(n)$	$O(1) to O(n)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• LeetCode provides 2 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: Iterate through all integers from 1 to num.

Hint 2: For any integer, extract the individual digits to compute their sum and check if it is even.