The problem titled “Subtract the Product and Sum of Digits of an Integer” is one of the simpler questions you might come across on Leetcode. Despite its simplicity, this problem offers a good opportunity to brush up on basic Python programming constructs and techniques. In this guide, we’ll dissect the problem and explore multiple solutions with varying degrees of efficiency.

## Table of Contents

- Problem Description
- Approach 1: Naive Approach
- Approach 2: Optimized Approach
- Approach 3: Pythonic Way
- Testing and Validation
- Time and Space Complexity
- Conclusion

## 1. Problem Description

The problem asks you to subtract the sum of digits of a given integer from the product of its digits. Specifically, given an integer number `n`

, you are required to return the difference between the product and sum of its digits.

### Example:

**Input**: `n = 234`

**Output**: `15`

**Explanation**:

Product of digits: 2×3×4=24

Sum of digits: 2+3+4=9

Difference: 24−9=15

## 2. Approach 1: Naive Approach

### Algorithm

- Convert the integer into a string.
- Split the string into individual digits.
- Calculate the product and sum of the digits.
- Return the difference between the product and sum.

### Python Code

```
def subtractProductAndSum(n):
str_n = str(n)
product = 1
sum_digits = 0
for digit in str_n:
int_digit = int(digit)
product *= int_digit
sum_digits += int_digit
return product - sum_digits
print(subtractProductAndSum(234)) # Output should be 15
```

## 3. Approach 2: Optimized Approach

### Algorithm

- Initialize variables for the product (
`product = 1`

) and sum (`sum_digits = 0`

). - Loop through the integer by continuously dividing it by 10, taking the remainder each time to get the last digit.
- Update the product and sum variables during each iteration.

### Python Code

```
def subtractProductAndSum(n):
product = 1
sum_digits = 0
while n > 0:
digit = n % 10
product *= digit
sum_digits += digit
n //= 10
return product - sum_digits
print(subtractProductAndSum(234)) # Output should be 15
```

## 4. Approach 3: Pythonic Way

Python offers various techniques that can be very useful for solving simple problems like this in a concise manner.

### Python Code

```
from functools import reduce
from operator import mul
def subtractProductAndSum(n):
digits = [int(d) for d in str(n)]
return reduce(mul, digits) - sum(digits)
print(subtractProductAndSum(234)) # Output should be 15
```

## 5. Testing and Validation

It’s crucial to test your solution on multiple test cases to make sure it works as expected. For this problem, edge cases are minimal due to the simplicity of the problem. A few test cases to consider:

- Single-digit integers
- Integers with the same repeated digits
- Large integers

## 6. Time and Space Complexity

**Approach 1**:- Time Complexity: O(d) where d is the number of digits in n.
- Space Complexity: O(d)

**Approach 2**:- Time Complexity: O(d)
- Space Complexity: O(1)

**Approach 3**:- Time Complexity: O(d)
- Space Complexity: O(d)

## 7. Conclusion

We explored three different ways to solve the “Subtract the Product and Sum of Digits of an Integer” problem from Leetcode. The first approach was a naive method, converting the integer to a string and iterating through it. The second approach optimized the space complexity by performing all operations on the integer itself. Finally, the third approach showcased Python’s concise syntax and built-in functions to solve the problem in a Pythonic manner.