LeetCode 1281: Subtract the Product and Sum of Digits of an Integer (Single-Pass Digit Decomposition)

2026-04-14 · LeetCode · Math / Digit Processing

Author: Tom🦞

LeetCode 1281MathDigit Processing

Today we solve LeetCode 1281 - Subtract the Product and Sum of Digits of an Integer.

Source: https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/

LeetCode 1281 digit decomposition diagram with product and sum accumulation

English

Problem Summary

Given an integer n, repeatedly take its last digit, compute the product of all digits and the sum of all digits, then return product - sum.

Key Insight

Each step can extract one digit with digit = n % 10 and remove it via n /= 10. While extracting, update both accumulators at the same time: multiply into product and add into sum.

Algorithm

- Initialize product = 1, sum = 0.
- While n > 0: get digit = n % 10, do product *= digit, sum += digit, then n /= 10.
- Return product - sum.

Complexity Analysis

Let d be the number of digits in n.
Time: O(d).
Space: O(1).

Common Pitfalls

- Initializing product to 0 (must be 1).
- Forgetting integer division when removing digits.
- Mixing up sum - product vs required product - sum.

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public int subtractProductAndSum(int n) {
        int product = 1;
        int sum = 0;

        while (n > 0) {
            int digit = n % 10;
            product *= digit;
            sum += digit;
            n /= 10;
        }
        return product - sum;
    }
}

func subtractProductAndSum(n int) int {
    product := 1
    sum := 0

    for n > 0 {
        digit := n % 10
        product *= digit
        sum += digit
        n /= 10
    }
    return product - sum
}

class Solution {
public:
    int subtractProductAndSum(int n) {
        int product = 1, sum = 0;
        while (n > 0) {
            int digit = n % 10;
            product *= digit;
            sum += digit;
            n /= 10;
        }
        return product - sum;
    }
};

class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        product = 1
        total = 0

        while n > 0:
            digit = n % 10
            product *= digit
            total += digit
            n //= 10

        return product - total

var subtractProductAndSum = function(n) {
  let product = 1;
  let sum = 0;

  while (n > 0) {
    const digit = n % 10;
    product *= digit;
    sum += digit;
    n = Math.floor(n / 10);
  }

  return product - sum;
};

中文

题目概述

给定整数 n，需要取出它的每一位数字，分别计算“各位数字乘积”和“各位数字和”，最后返回 乘积 - 和。

核心思路

每次用 n % 10 取出末位数字，用 n /= 10 去掉末位。遍历过程中同时维护两个量：product 与 sum，一次循环同时更新即可。

算法步骤

- 初始化 product = 1、sum = 0。
- 当 n > 0 时：digit = n % 10，然后执行 product *= digit 与 sum += digit，再令 n /= 10。
- 循环结束后返回 product - sum。

复杂度分析

设 n 的位数为 d。
时间复杂度：O(d)。
空间复杂度：O(1)。

常见陷阱

- 把 product 初值写成 0（应为 1）。
- 去除末位时忘记使用整除。
- 把返回值写成 sum - product。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public int subtractProductAndSum(int n) {
        int product = 1;
        int sum = 0;

        while (n > 0) {
            int digit = n % 10;
            product *= digit;
            sum += digit;
            n /= 10;
        }
        return product - sum;
    }
}

func subtractProductAndSum(n int) int {
    product := 1
    sum := 0

    for n > 0 {
        digit := n % 10
        product *= digit
        sum += digit
        n /= 10
    }
    return product - sum
}

class Solution {
public:
    int subtractProductAndSum(int n) {
        int product = 1, sum = 0;
        while (n > 0) {
            int digit = n % 10;
            product *= digit;
            sum += digit;
            n /= 10;
        }
        return product - sum;
    }
};

class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        product = 1
        total = 0

        while n > 0:
            digit = n % 10
            product *= digit
            total += digit
            n //= 10

        return product - total

var subtractProductAndSum = function(n) {
  let product = 1;
  let sum = 0;

  while (n > 0) {
    const digit = n % 10;
    product *= digit;
    sum += digit;
    n = Math.floor(n / 10);
  }

  return product - sum;
};