What is the most efficient way to subtract the product and sum of digits of an integer?

Iterate through each digit using modulo and division, maintain a running product and sum, and subtract sum from product at the end.

Can I convert the integer to a string to solve this problem?

Yes, converting to a string is valid, but direct modulo/division iteration is more memory efficient and aligns with the math-driven pattern.

What are common mistakes when solving this problem?

Common mistakes include initializing the product as 0, skipping digits in the loop, or miscalculating the subtraction order.

Does this approach work for the maximum constraint n = 100000?

Yes, iterating through at most 6 digits is efficient, keeping both time and space complexity minimal.

Why is this called a math-driven solution strategy?

Because the solution relies on fundamental digit arithmetic—product and sum operations—rather than data structures or advanced algorithms.

#1281

Easy

auto_awesome数学·driven

LeetCode 题解工作台

整数的各位积和之差

给你一个整数 n ，请你帮忙计算并返回该整数「各位数字之积」与「各位数字之和」的差。示例 1：输入： n = 234 输出： 15 解释：各位数之积 = 2 * 3 * 4 = 24 各位数之和 = 2 + 3 + 4 = 9 结果 = 24 - 9 = 15 示例 2：输入： n = 44…

数学

题目描述

给你一个整数 n，请你帮忙计算并返回该整数「各位数字之积」与「各位数字之和」的差。

示例 1：

输入：n = 234
输出：15 
解释：
各位数之积 = 2 * 3 * 4 = 24 
各位数之和 = 2 + 3 + 4 = 9 
结果 = 24 - 9 = 15

示例 2：

输入：n = 4421
输出：21
解释： 
各位数之积 = 4 * 4 * 2 * 1 = 32 
各位数之和 = 4 + 4 + 2 + 1 = 11 
结果 = 32 - 11 = 21

提示：

1 <= n <= 10^5

lightbulb

解题思路

方法一：模拟

我们用两个变量 $x$ 和 $y$ 分别记录各位数之积、各位数之和，初始时 $x=1,y=0$ 。

当 $n \gt 0$ 时，每次将 $n$ 对 $10$ 取模得到当前位的数字 $v$ ，并将 $n$ 除以 $10$ 后继续进行下一次循环。在每次循环中，我们更新 $x = x \times v$ , $y = y + v$ 。

最终，我们返回 $x - y$ 即可。

时间复杂度 $O(\log n)$ ，其中 $n$ 是题目给定的整数。空间复杂度 $O(1)$ 。

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class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        nums = list(map(int, str(n)))
        return prod(nums) - sum(nums)

speed

复杂度分析

指标	值
时间	complexity is O(d) where d is the number of digits in n, since each digit is processed once. Space complexity is O(1) because only two integer variables are used regardless of n's size.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Ask candidates to handle each digit efficiently without converting to arrays unnecessarily.
question_mark
Check if the solution correctly initializes the product to 1 to avoid zeroing out the result.
question_mark
Probe understanding of basic number manipulation and iteration over digits in an integer.

warning

常见陷阱

外企场景

error
Initializing the product incorrectly as 0 instead of 1, causing the final result to always be zero.
error
Skipping digits when using division and modulo incorrectly in the loop.
error
Using excessive extra space instead of simple integer tracking for product and sum.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Compute the difference for very large integers requiring optimized iteration or string handling.
arrow_right_alt
Return both product and sum separately instead of their difference for additional insight into digit composition.
arrow_right_alt
Handle negative integers by considering absolute values or separate sign handling.

help

常见问题

外企场景

继续练习

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