Solution Explanation: Subtract the Product and Sum of Digits of an Integer

This problem involves calculating the difference between the product and the sum of the digits of a given integer. The solution employs a straightforward iterative approach.

Approach:

1. Initialization: We initialize two variables: product (to store the product of digits, starting at 1 to handle multiplication) and sum (to store the sum of digits, starting at 0).

2. Iteration: We iterate through the digits of the integer. This is efficiently done using a loop and the modulo operator (%) along with integer division (/).

   • In each iteration:
     • We extract the last digit using the modulo operator (n % 10).
     • We update the product by multiplying it with the extracted digit.
     • We update the sum by adding the extracted digit.
     • We remove the last digit from the number using integer division (n /= 10 or n = n / 10).

3. Return Value: Finally, we return the difference between the product and the sum.

Time Complexity Analysis:

The time complexity is O(log₁₀ n). The number of iterations in the loop is proportional to the number of digits in the input integer n. The number of digits in n is approximately log₁₀ n. Therefore, the algorithm's runtime grows logarithmically with the input size.

Space Complexity Analysis:

The space complexity is O(1). We use a constant number of variables regardless of the size of the input integer.

Code Examples in Multiple Languages:

Python:

def subtractProductAndSum(n: int) -> int:
    product = 1
    sum_digits = 0
    while n > 0:
        digit = n % 10
        product *= digit
        sum_digits += digit
        n //= 10  # Integer division
    return product - sum_digits
 

Java:

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;
    }
}

C++:

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;
    }
};

JavaScript:

const subtractProductAndSum = (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;
};

Go:

func subtractProductAndSum(n int) int {
    product := 1
    sum := 0
    for n > 0 {
        digit := n % 10
        product *= digit
        sum += digit
        n /= 10
    }
    return product - sum
}

These examples all follow the same algorithmic approach, differing only in syntax specific to each programming language. The core logic remains consistent across all implementations.