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.

#2180

Easy

auto_awesome数学·结合·模拟

LeetCode 题解工作台

统计各位数字之和为偶数的整数个数

给你一个正整数 num ，请你统计并返回小于或等于 num 且各位数字之和为偶数的正整数的数目。正整数的各位数字之和是其所有位上的对应数字相加的结果。示例 1：输入： num = 4 输出： 2 解释：只有 2 和 4 满足小于等于 4 且各位数字之和为偶数。示例 2：输入： …

数学模拟

题目描述

给你一个正整数 num ，请你统计并返回 小于或等于 num 且各位数字之和为偶数的正整数的数目。

正整数的 各位数字之和 是其所有位上的对应数字相加的结果。

示例 1：

输入：num = 4
输出：2
解释：
只有 2 和 4 满足小于等于 4 且各位数字之和为偶数。

示例 2：

输入：num = 30
输出：14
解释：
只有 14 个整数满足小于等于 30 且各位数字之和为偶数，分别是： 
2、4、6、8、11、13、15、17、19、20、22、24、26 和 28 。

提示：

1 <= num <= 1000

lightbulb

解题思路

方法一：枚举

一种最简单直接的方法是枚举 $[1,..num]$ 的所有整数 $x$ ，判断 $x$ 各位数字之和是否为偶数，是则答案加一。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(1)$ 。其中 $n$ 为 $num$ 的值。

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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

speed

复杂度分析

指标	值
时间	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.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

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