Subtract the Product and Sum of Digits of an Integer

EASY

Description

Given an integer number 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 <= 10^5

Approaches

Checkout 2 different approaches to solve Subtract the Product and Sum of Digits of an Integer. Click on different approaches to view the approach and algorithm in detail.

String Conversion and Iteration

This approach involves converting the integer into a string. Once we have the string representation, we can iterate through each character, convert it back to its numeric value, and then calculate the product and sum of these digits.

Algorithm

Convert the input integer n to its string representation.
Initialize two variables: sumOfDigits to 0 and productOfDigits to 1.
Iterate through each character of the string.
For each character, convert it to its integer equivalent.
Add this integer value to sumOfDigits.
Multiply productOfDigits by this integer value.
After the loop finishes, return the difference productOfDigits - sumOfDigits.

The core idea is to leverage built-in string conversion functions. We first convert the number n into a string. Then, we can easily iterate over this string. In each iteration, we get a character representing a digit, which we convert back to an integer. We maintain two running variables, one for the sum and one for the product, updating them with each digit. Finally, we compute the difference.

class Solution {
    public int subtractProductAndSum(int n) {
        String s = Integer.toString(n);
        int sumOfDigits = 0;
        int productOfDigits = 1;

        for (char c : s.toCharArray()) {
            int digit = Character.getNumericValue(c);
            sumOfDigits += digit;
            productOfDigits *= digit;
        }

        return productOfDigits - sumOfDigits;
    }
}

Complexity Analysis

Time Complexity: O(log10(n)). The time taken is proportional to the number of digits in `n`. Converting an integer to a string takes O(log10(n)) time, and iterating through the string also takes O(log10(n)) time.Space Complexity: O(log10(n)). We need extra space to store the string representation of the number, and the length of the string is proportional to the number of digits in `n`.

Pros and Cons

Pros:

Conceptually simple and easy to read for those familiar with string manipulation.

Cons:

Less efficient in terms of space compared to the mathematical approach due to the creation of an intermediate string.
May be slightly slower in practice due to the overhead of string conversion and character parsing.

Code Solutions

Checking out 4 solutions in different languages for Subtract the Product and Sum of Digits of an Integer. Click on different languages to view the code.

public class Solution {
    public int SubtractProductAndSum(int n) {
        int x = 1;
        int y = 0;
        for (; n > 0; n /= 10) {
            int v = n % 10;
            x *= v;
            y += v;
        }
        return x - y;
    }
}

Video Solution

Watch the video walkthrough for Subtract the Product and Sum of Digits of an Integer

Patterns:

Math

Companies:

Quora |