What is the key approach to solving the Difference Between Element Sum and Digit Sum of an Array?

The key approach is to compute both the element sum (the sum of all numbers) and the digit sum (the sum of all digits of each number), and then return the absolute difference.

What are common pitfalls when solving this problem?

Common pitfalls include forgetting to calculate the digit sum correctly, not handling the absolute value properly, and missing edge cases like arrays with single digits.

How can I optimize my solution for large arrays in this problem?

You can optimize by using a simple for loop to calculate both sums in one pass over the array. Additionally, avoid unnecessary space usage by performing calculations on the fly.

What time complexity should I expect when solving this problem?

The time complexity is O(n), where n is the length of the array, since you need to process each element and calculate its digit sum individually.

Is there a more efficient way to compute the digit sum?

A common way to compute the digit sum is to repeatedly extract the last digit using modulo 10 and then reduce the number by dividing it by 10.

#2535

Easy

auto_awesome数组·数学

LeetCode 题解工作台

数组元素和与数字和的绝对差

给你一个正整数数组 nums 。元素和是 nums 中的所有元素相加求和。数字和是 nums 中每一个元素的每一数位（重复数位需多次求和）相加求和。返回元素和与数字和的绝对差。注意：两个整数 x 和 y 的绝对差定义为 |x - y| 。示例 1：输入： nums = [1…

数组数学

题目描述

给你一个正整数数组 nums 。

元素和 是 nums 中的所有元素相加求和。
数字和 是 nums 中每一个元素的每一数位（重复数位需多次求和）相加求和。

返回 元素和 与 数字和 的绝对差。

注意：两个整数 x 和 y 的绝对差定义为 |x - y| 。

示例 1：

输入：nums = [1,15,6,3]
输出：9
解释：
nums 的元素和是 1 + 15 + 6 + 3 = 25 。
nums 的数字和是 1 + 1 + 5 + 6 + 3 = 16 。
元素和与数字和的绝对差是 |25 - 16| = 9 。

示例 2：

输入：nums = [1,2,3,4]
输出：0
解释：
nums 的元素和是 1 + 2 + 3 + 4 = 10 。
nums 的数字和是 1 + 2 + 3 + 4 = 10 。
元素和与数字和的绝对差是 |10 - 10| = 0 。

提示：

1 <= nums.length <= 2000
1 <= nums[i] <= 2000

lightbulb

解题思路

方法一：模拟

我们遍历数组 $\textit{nums}$ ，计算元素和 $x$ 和数字和 $y$ ，最后返回 $|x - y|$ 即可。由于 $x$ 一定大于等于 $y$ ，所以我们也可以直接返回 $x - y$ 。

时间复杂度 $O(n \times \log_{10} M)$ ，其中 $n$ 和 $M$ 分别是数组 $\textit{nums}$ 的长度和数组中元素的最大值。空间复杂度 $O(1)$ 。

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class Solution:
    def differenceOfSum(self, nums: List[int]) -> int:
        x = y = 0
        for v in nums:
            x += v
            while v:
                y += v % 10
                v //= 10
        return x - y

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Tests ability to loop through an array efficiently.
question_mark
Checks understanding of digit extraction from integers.
question_mark
Evaluates correctness in calculating absolute differences.

warning

常见陷阱

外企场景

error
Forgetting to correctly sum the digits of numbers with multiple digits.
error
Not taking the absolute value of the difference between sums.
error
Incorrectly handling large numbers or edge cases with single-digit arrays.

swap_horiz

进阶变体

外企场景

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Change the array to contain only single-digit numbers.
arrow_right_alt
Increase the array size to test scalability.
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Introduce negative integers to test how absolute values are handled.

help

常见问题

外企场景

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