Check Divisibility by Digit Sum and Product - Video Solutions

Problem Overview: Given an integer n, compute the sum and product of its digits. The task is to determine whether n is divisible by the value formed from those digits (commonly digitSum + digitProduct). The problem focuses on basic number manipulation using simple math operations.

Approach 1: Digit Simulation (O(d) time, O(1) space)

Iterate through every digit of the number and accumulate two values: the sum of digits and the product of digits. You can extract digits using repeated modulo and division operations (digit = n % 10, then n //= 10). After processing all digits, compute the final divisor (typically digitSum + digitProduct) and check whether the original number is divisible by it using a modulo operation.

This approach relies purely on arithmetic digit extraction, which avoids extra memory allocations. The runtime depends on the number of digits d, giving O(d) time complexity and constant O(1) space. This pattern appears frequently in math and simulation problems where you repeatedly process digits of an integer.

Approach 2: String-Based Digit Traversal (O(d) time, O(d) space)

Convert the integer to a string and iterate through each character. Convert each character back to an integer digit, update the running digit sum and digit product, and compute the same divisibility condition at the end. The logic is straightforward and often easier to read in high-level languages.

This version still runs in O(d) time because every digit is processed once. However, converting the number to a string requires O(d) extra space. Many developers prefer this version during interviews when readability matters more than strict memory usage.

Recommended for interviews: The arithmetic simulation approach. It demonstrates comfort with digit manipulation and avoids unnecessary allocations. Showing both methods helps: the string version proves clarity of thought, while the arithmetic method shows deeper control over numeric operations and constant-space optimization.