Subtract the Product and Sum of Digits of an Integer

Subtract the Product and Sum of Digits of an Integer - Problem

Math Easy

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

The product is calculated by multiplying all digits together, and the sum is calculated by adding all digits together.

For example, if n = 234, the product is 2 × 3 × 4 = 24 and the sum is 2 + 3 + 4 = 9, so the answer is 24 - 9 = 15.

Input & Output

Example 1 — Basic Three Digit Number

$ Input: n = 234

› Output: 15

💡 Note: Product of digits: 2 × 3 × 4 = 24. Sum of digits: 2 + 3 + 4 = 9. Difference: 24 - 9 = 15

Example 2 — Single Digit Number

$ Input: n = 4

› Output: 0

💡 Note: Product of digits: 4. Sum of digits: 4. Difference: 4 - 4 = 0

Example 3 — Number with Zero

$ Input: n = 4021

› Output: -7

💡 Note: Product of digits: 4 × 0 × 2 × 1 = 0. Sum of digits: 4 + 0 + 2 + 1 = 7. Difference: 0 - 7 = -7

Constraints

1 ≤ n ≤ 10⁵

Visualization

Tap to expand

Understanding the Visualization

Input

Integer n with multiple digits

Process

Calculate product and sum of all digits

Output

Return product - sum

Key Takeaway

🎯 Key Insight: Use modulo operator to extract digits and process both operations in a single pass

Asked in

a Amazon 12 M Microsoft 8

The key insight is to process each digit exactly once using modulo and division operations. Extract digits from right to left using n % 10 and n /= 10, maintaining running product and sum. Best approach is single pass calculation with Time: O(log n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Extract Digits One by One	O(log n)	O(1)	Extract each digit using modulo and division, calculate product and sum separately
Single Pass Calculation	O(log n)	O(1)	Calculate product and sum in one pass through the digits

Extract Digits One by One — Algorithm Steps

Step 1: Initialize product = 1 and sum = 0
Step 2: Extract each digit using n % 10, update product and sum
Step 3: Remove last digit using n = n // 10, repeat until n becomes 0
Step 4: Return product - sum

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Set product = 1, sum = 0

Extract & Calculate

Get each digit, update both values

Final Result

Return product - sum

Code -

solution.c — C

#include <stdio.h>

int solution(int n) {
    int product = 1;
    int sumDigits = 0;
    
    while (n > 0) {
        int digit = n % 10;
        product *= digit;
        sumDigits += digit;
        n /= 10;
    }
    
    return product - sumDigits;
}

int main() {
    int n;
    scanf("%d", &n);
    
    int result = solution(n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We process each digit once, and there are log₁₀(n) digits in number n

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables for product and sum

✓ Linear Space

25.0K Views

Medium Frequency

~5 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Extract Digits One by One — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler