Watch 10 video solutions for Subtract the Product and Sum of Digits of an Integer, a easy level problem involving Math. This walkthrough by Fisher Coder has 3,492 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
n, return the difference between the product of its digits and the sum of its digits.
Example 1:
Input: n = 234 Output: 15 Explanation: Product of digits = 2 * 3 * 4 = 24 Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 15
Example 2:
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 <= 10^5Problem Overview: You are given an integer n. Compute the product of its digits and the sum of its digits, then return product - sum. The task mainly tests basic math operations and how you extract digits from a number.
Approach 1: Loop Through Digits (O(d) time, O(1) space)
This is the standard mathematical approach. Repeatedly extract the last digit using n % 10, update both the running product and running sum, and then remove the digit with integer division n // 10. Continue until the number becomes zero. Each iteration processes one digit, so the total work is proportional to the number of digits d.
The key idea is that modulus gives the current digit and division shifts the number right by one decimal place. This avoids converting the number to another data type and keeps memory usage constant. The algorithm performs simple arithmetic operations while iterating through digits, making it both efficient and easy to reason about. Time complexity is O(d) and space complexity is O(1).
This pattern appears frequently in math and simulation problems where you need to process digits individually.
Approach 2: String Conversion (O(d) time, O(d) space)
Another option converts the integer into a string and iterates over its characters. For each character, convert it back to a digit and update the product and sum variables. This method removes the need for modulus and division operations, which some developers find easier to read.
The logic becomes straightforward: iterate through the string, compute digit = ch - '0' (or equivalent conversion), multiply it into the product, and add it to the sum. Because the string representation stores all digits, this approach requires O(d) additional space. Time complexity remains O(d) since every digit is processed once.
This method is often used when working with string manipulation or when consistency with other string-based operations matters.
Recommended for interviews: The loop-through-digits math approach is what interviewers usually expect. It demonstrates comfort with digit extraction using modulus and division, a common pattern in integer manipulation problems. The string approach still works and is readable, but it adds unnecessary space usage. Showing the math-based method signals stronger problem-solving fundamentals.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Loop Through Digits (Math) | O(d) | O(1) | Best general solution; minimal memory and standard interview expectation |
| String Conversion | O(d) | O(d) | Useful when already processing digits as strings or prioritizing readability |