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Subtract the Product and Sum of Digits of an Integer

Easy

Quora

2 views

Topics:

Math

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

Try coding it in LeetCode

Subtract the Product and Sum of Digits of an Integer

Easy

Quora

2 views

Topics:

Math

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer 'n'? Specifically, what are the minimum and maximum possible values?
Is the input integer 'n' guaranteed to be non-negative?
Should I handle the case where 'n' is zero? If so, what should the return value be?
Are there any limitations on the size of the integer datatype, and should I be concerned about integer overflow during the product calculation?
Are there any specific error conditions that I should handle, or can I assume the input will always be a valid integer?

Brute Force Solution

Approach

We need to find the product and sum of the digits of a given number, and then subtract the sum from the product. The brute force way means we'll directly calculate the product and sum by looking at each digit one by one.

Here's how the algorithm would work step-by-step:

First, take the given number.
Isolate each digit of the number, one at a time, working from right to left.
Keep a running total of the sum of all the digits you've isolated.
Also, keep a running total of the product of all the digits you've isolated.
Once you've gone through all the digits, subtract the final sum from the final product.
The result of the subtraction is the answer.

Code Implementation

def subtract_product_and_sum(number):
    product_of_digits = 1
    sum_of_digits = 0

    number_copy = number # Preserving the original input

    while number_copy > 0:
        # Extract the last digit
        last_digit = number_copy % 10

        # Update the product
        product_of_digits *= last_digit

        # Update the sum
        sum_of_digits += last_digit

        # Remove the last digit from the number
        number_copy //= 10

    return product_of_digits - sum_of_digits

Big(O) Analysis

Time Complexity

O(log n) – The algorithm iterates through the digits of the input number n. The number of digits in a number n is proportional to log n (base 10). Therefore, the time complexity of extracting each digit and calculating the product and sum is O(log n). The final subtraction is a constant time operation O(1) and doesn't affect the overall complexity.

Space Complexity

O(1) – The algorithm maintains two integer variables, one for the running sum of digits and another for the running product of digits. These variables consume a fixed amount of memory regardless of the number of digits in the input integer N. Thus, the auxiliary space required remains constant. Since there are no other additional data structures used, the space complexity is O(1).

Optimal Solution

Approach

To find the difference between the product and sum of digits, we need to calculate the product and sum separately. We can do this by looking at each digit of the number one at a time and updating our running product and sum.

Here's how the algorithm would work step-by-step:

Take the number and look at one digit at a time, from right to left.
Multiply each digit with a running product to get the product of all digits.
Add each digit to a running sum to get the sum of all digits.
Once you've processed all the digits, subtract the sum from the product.
The result of this subtraction is the answer.

Code Implementation

def subtract_product_and_sum(number):
    product_of_digits = 1
    sum_of_digits = 0

    number_string = str(number)

    for digit_char in number_string:
        digit = int(digit_char)
        # Accumulate product; handles edge case of 0
        product_of_digits *= digit

        # Accumulate sum of digits
        sum_of_digits += digit

    # Calculate final result.
    result = product_of_digits - sum_of_digits
    return result

Big(O) Analysis

Time Complexity

O(log(n)) – The time complexity is determined by the number of digits in the input integer 'n'. The algorithm iterates through each digit of the number. The number of digits in an integer 'n' is proportional to log(n) (base 10). Therefore, the loop runs for log(n) iterations to extract and process each digit. Consequently, the overall time complexity is O(log(n)).

Space Complexity

O(1) – The provided algorithm calculates the product and sum of digits using only a few variables to store the running product, running sum, and the current digit being processed. The number of these variables remains constant regardless of the input integer's size, N (the number of digits). Therefore, no auxiliary data structures scale with the input, and the space complexity is constant.

Edge Cases

Case	How to Handle
Input is zero	The product of digits will be 0 and the sum will be 0, so the result will be 0.
Input is a single-digit number	The product and sum will be the number itself, resulting in a difference of 0.
Integer overflow during product calculation with large numbers	Use a larger data type (e.g., long) for the product to prevent overflow, or check for overflow during calculation and return an error code.
Negative input	Take the absolute value of the input number as the sign does not impact individual digits' product or sum.
Input consists of many 1s	The product will be 1, and the sum will be the number of digits.
Input consists of many 0s, besides a single non-zero digit	The product will be 0, and the sum will be the non-zero digit.
Maximum integer value (Integer.MAX_VALUE)	Handle potential overflow when calculating the product and sum of its digits.
All digits are large (e.g., nines)	Consider the impact of multiple large digits when calculating the product and sum and handle potential overflows.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer 'n'? Specifically, what are the minimum and maximum possible values?
Is the input integer 'n' guaranteed to be non-negative?
Should I handle the case where 'n' is zero? If so, what should the return value be?
Are there any limitations on the size of the integer datatype, and should I be concerned about integer overflow during the product calculation?
Are there any specific error conditions that I should handle, or can I assume the input will always be a valid integer?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, take the given number.
Isolate each digit of the number, one at a time, working from right to left.
Keep a running total of the sum of all the digits you've isolated.
Also, keep a running total of the product of all the digits you've isolated.
Once you've gone through all the digits, subtract the final sum from the final product.
The result of the subtraction is the answer.

Code Implementation

def subtract_product_and_sum(number):
    product_of_digits = 1
    sum_of_digits = 0

    number_copy = number # Preserving the original input

    while number_copy > 0:
        # Extract the last digit
        last_digit = number_copy % 10

        # Update the product
        product_of_digits *= last_digit

        # Update the sum
        sum_of_digits += last_digit

        # Remove the last digit from the number
        number_copy //= 10

    return product_of_digits - sum_of_digits

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Take the number and look at one digit at a time, from right to left.
Multiply each digit with a running product to get the product of all digits.
Add each digit to a running sum to get the sum of all digits.
Once you've processed all the digits, subtract the sum from the product.
The result of this subtraction is the answer.

Code Implementation

def subtract_product_and_sum(number):
    product_of_digits = 1
    sum_of_digits = 0

    number_string = str(number)

    for digit_char in number_string:
        digit = int(digit_char)
        # Accumulate product; handles edge case of 0
        product_of_digits *= digit

        # Accumulate sum of digits
        sum_of_digits += digit

    # Calculate final result.
    result = product_of_digits - sum_of_digits
    return result

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input is zero	The product of digits will be 0 and the sum will be 0, so the result will be 0.
Input is a single-digit number	The product and sum will be the number itself, resulting in a difference of 0.
Integer overflow during product calculation with large numbers	Use a larger data type (e.g., long) for the product to prevent overflow, or check for overflow during calculation and return an error code.
Negative input	Take the absolute value of the input number as the sign does not impact individual digits' product or sum.
Input consists of many 1s	The product will be 1, and the sum will be the number of digits.
Input consists of many 0s, besides a single non-zero digit	The product will be 0, and the sum will be the non-zero digit.
Maximum integer value (Integer.MAX_VALUE)	Handle potential overflow when calculating the product and sum of its digits.
All digits are large (e.g., nines)	Consider the impact of multiple large digits when calculating the product and sum and handle potential overflows.