#2180

Easy

LeetCode Problem Workspace

Count Integers With Even Digit Sum

Solve this Easy Math plus Simulation problem by counting numbers whose digit sums are even up to a given limit efficiently.

math simulation

Problem Statement

Given a positive integer num, determine how many positive integers less than or equal to num have digit sums that are even. The digit sum is the total of all digits in a number.

For example, if num = 30, count all numbers from 1 to 30 where the sum of digits is even. Return the total count as the result.

Examples

Example 1

Input: num = 4

Output: 2

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

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

Solution Approach

Iterate Through All Numbers

Loop from 1 to num, compute the digit sum for each number, and check if it is even. Increment a counter for every number that satisfies this condition.

Calculate Digit Sum Efficiently

Convert the number to a string or repeatedly divide by 10 to extract digits. Sum the digits directly instead of using unnecessary data structures.

Return the Count

After iterating through all numbers, return the counter as the total number of integers with even digit sums. This finalizes the simulation approach cleanly.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(num * d) where d is the number of digits, due to computing digit sums for each number. Space complexity is O(1) since only a counter and temporary variables are used.

What Interviewers Usually Probe

Check if the candidate immediately recognizes the Math plus Simulation pattern.
Listen for an explanation of digit sum computation without extra arrays.
Watch if they avoid overcomplicating the iteration with unnecessary data structures.

Common Pitfalls or Variants

Common pitfalls

Forgetting that single-digit numbers still count if their digit sum is even.
Using string conversion without considering efficiency in larger num ranges.
Returning an incorrect count by checking for odd sums instead of even sums.

Follow-up variants

Count numbers with odd digit sums instead of even.
Return a list of all numbers with even digit sums rather than just the count.
Handle larger num values efficiently using modulo arithmetic patterns.

FAQ

What is the main pattern for Count Integers With Even Digit Sum?

The primary pattern is Math plus Simulation: iterating through numbers and calculating digit sums to check parity.

Can this problem be solved without converting numbers to strings?

Yes, repeatedly dividing by 10 and summing remainders avoids string conversion and keeps space complexity minimal.

What is the expected time complexity?

O(num * d), where d is the number of digits in each number, because each digit must be summed once per number.

Are there common mistakes to avoid?

Yes, including counting numbers with odd sums, ignoring single-digit numbers, or miscomputing the digit sum loop.

Does GhostInterview provide example explanations?

Yes, it shows worked examples like num = 4 or num = 30 to clarify which numbers contribute to the even digit sum count.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def countEven(self, num: int) -> int:
        ans = 0
        for x in range(1, num + 1):
            s = 0
            while x:
                s += x % 10
                x //= 10
            ans += s % 2 == 0
        return ans

Solution 2

#### Python3

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class Solution:
    def countEven(self, num: int) -> int:
        ans = 0
        for x in range(1, num + 1):
            s = 0
            while x:
                s += x % 10
                x //= 10
            ans += s % 2 == 0
        return ans

Continue Practicing

#2177 find three consecutive integers that sum to a given number #2169 count operations to obtain zero #2221 find triangular sum of an array