Subtract the Product and Sum of Digits of an Integer

Difficulty: Easy | Acceptance: 86.60% | Paid: No Topics: Math

Given an integer number n, return the difference between the product of its digits and the sum of its digits.

• Examples
• Constraints
• Approach 1: Iterative Digit Extraction
• Approach 2: String Conversion
• Approach 3: Functional Reduction

Examples

Input: n = 234
Output: 15
Explanation:
Product of digits = 2 * 3 * 4 = 24
Sum of digits = 2 + 3 + 4 = 9
Result = 24 - 9 = 15

Input: n = 4421
Output: 21
Explanation:
Product of digits = 4 * 4 * 2 * 1 = 32
Sum of digits = 4 + 4 + 2 + 1 = 11
Result = 32 - 11 = 21

Constraints

1 <= n <= 10^5

Iterative Digit Extraction

Intuition Extract each digit using modulo and division operations, then compute product and sum separately.

Steps

• Initialize product to 1 and sum to 0
• While n is greater than 0, extract the last digit using modulo 10
• Multiply product by the digit and add digit to sum
• Remove the last digit by dividing n by 10
• Return the difference between product and sum

class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        product = 1
        sum_digits = 0
        while n > 0:
            digit = n % 10
            product *= digit
            sum_digits += digit
            n //= 10
        return product - sum_digits

Complexity

• Time: O(log n) - number of digits in n
• Space: O(1) - only using constant extra space
• Notes: Most efficient approach with minimal memory usage

String Conversion

Intuition Convert the number to a string to easily iterate over each digit character.

Steps

• Convert n to a string representation
• Iterate through each character in the string
• Convert each character back to an integer
• Calculate product and sum of all digits
• Return the difference

class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        digits = [int(d) for d in str(n)]
        product = 1
        sum_digits = 0
        for d in digits:
            product *= d
            sum_digits += d
        return product - sum_digits

Complexity

• Time: O(log n) - number of digits in n
• Space: O(log n) - string representation of n
• Notes: More readable but uses extra memory for string conversion

Functional Reduction

Intuition Use functional programming constructs like reduce to compute product and sum in a declarative way.

Steps

• Convert n to string and split into individual digit characters
• Map each character to its integer value
• Use reduce to compute product with initial value 1
• Use reduce or sum to compute total of all digits
• Return the difference

from functools import reduce
import operator

class Solution:
    def subtractProductAndSum(self, n: int) -> int:
        digits = [int(d) for d in str(n)]
        product = reduce(operator.mul, digits, 1)
        sum_digits = sum(digits)
        return product - sum_digits

Complexity

• Time: O(log n) - number of digits in n
• Space: O(log n) - array of digits
• Notes: Clean functional style but with overhead of array allocation