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Given an integer number n, return the difference between the product of its digits and the sum of its digits.

Clarifying Questions:

  1. Input Range: Are there any constraints on the input number n (e.g., the range of n)?
    • Generally, we assume n to be a non-negative integer within the typical integer range.
  2. Edge Cases: How should we handle cases such as 0 or single-digit numbers?
    • For 0, the product and sum will both be 0, so the result should be 0.
    • For single-digit values, the product and sum will be the same, resulting in a difference of 0.

No additional clarifications seem necessary. Let’s proceed with the solution.

Strategy:

  1. Extract Digits: Convert the integer n to a string to easily iterate through each digit.
  2. Initialize Variables: Initialize two variables to compute the product and sum of the digits.
  3. Compute Product and Sum: Traverse each digit, converting it back to an integer, and update the product and sum accordingly.
  4. Calculate Difference: Compute the difference between the product and the sum.
  5. Return Result: Return the computed difference.

Code:

def subtractProductAndSum(n: int) -> int:
    digits = [int(digit) for digit in str(n)]  # Convert number to list of digits
    product = 1
    total_sum = 0
    
    for digit in digits:
        product *= digit
        total_sum += digit
    
    return product - total_sum

# Example usage
print(subtractProductAndSum(234))  # Output: 15 (2*3*4 - (2+3+4) = 24 - 9 = 15)
print(subtractProductAndSum(4421))  # Output: 21 (4*4*2*1 - (4+4+2+1) = 32 - 11 = 21)

Time Complexity:

This completes the solution, ensuring efficiency and correctness based on the problem constraints and requirements.

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