What is the time complexity of Subtract the Product and Sum of Digits of an Integer?

The optimal solution has a time complexity of O(log(n)).

What data structures are used to solve Subtract the Product and Sum of Digits of an Integer?

This problem uses Math. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Subtract the Product and Sum of Digits of an Integer?

Leveraging string conversion (`str(n)`) is an elegant and readable way to iterate through the individual digits of an integer, effectively abstracting away more traditional modulo and division operations which can sometimes be more verbose. Correctly initializing accumulators is vital: the product accumulator must start at `1` to ensure that initial multiplications don't result in an incorrect zero-out, while the sum accumulator correctly starts at `0`.

Which companies ask LeetCode 1281. Subtract the Product and Sum of Digits of an Integer in interviews?

This problem is frequently asked at Quora.

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Subtract the Product and Sum of Digits of an Integer - LeetCode 1281 Solution

Q: How to solve Subtract the Product and Sum of Digits of an Integer?

The provided solution takes a very pragmatic and often Pythonic approach to digit extraction: it converts the integer `n` into a string. This is a brilliant maneuver because iterating over the characters of a string is inherently simple and direct. For `n = 234`, `str(n)` yields the string `'234'`. Now, each character, like `'2'`, `'3'`, and `'4'`, can be directly accessed in a loop. Inside the loop, each character (which is a string representation of a digit) is converted back into an integer using `int()`, making it ready for arithmetic operations. Simultaneously, the solution maintains two accumulator variables: `product` initialized to `1` and `sum_digits` initialized to `0`. Initializing `product` to `1` is crucial because multiplying by `0` would incorrectly result in `0` regardless of the other digits (unless `n` itself contained a 0, which would correctly make the product 0). As each digit is extracted and converted to an integer, it's immediately multiplied into `product` and added to `sum_digits`. This single-pass accumulation is highly efficient as it avoids iterating over the digits multiple times. Once the loop finishes processing all digits, both `product` and `sum_digits` hold their respective final values, and the last step is simply returning their difference.

Subtract the Product and Sum of Digits of an Integer - Complete Solution Guide

Subtract the Product and Sum of Digits of an Integer is LeetCode problem 1281, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

At its core, this problem asks us to perform two distinct aggregations on the individual digits of a given integer `n`: calculate their product and calculate their sum. Once we have these two results, the final step is a simple subtraction: `product - sum`. For instance, with `n = 234`, we first need to break it down into its constituent digits: 2, 3, and 4. Then, we compute their product (2 * 3 * 4 = 24) and their sum (2 + 3 + 4 = 9). The ultimate answer is the difference, 24 - 9 = 15. The interesting part here isn't the final subtraction, but rather the initial decomposition of the integer. How do we reliably get each digit out of a number like `n = 4421`? We need a robust way to access the '4', then the '4' again, then '2', then '1'. If `n` were a string, this would be trivial, but it's an integer. This requirement to deconstruct the number is the primary challenge to address before any arithmetic can begin. The constraints `1 <= n <= 10^5` tell us that `n` will always be positive and will have at most 6 digits (since 10^5 is 100,000, a 6-digit number). This means we don't have to worry about negative inputs, and the product of digits won't grow so large as to cause standard integer overflows in most languages, simplifying our considerations for variable types.

Solution Approach

The provided solution takes a very pragmatic and often Pythonic approach to digit extraction: it converts the integer `n` into a string. This is a brilliant maneuver because iterating over the characters of a string is inherently simple and direct. For `n = 234`, `str(n)` yields the string `'234'`. Now, each character, like `'2'`, `'3'`, and `'4'`, can be directly accessed in a loop. Inside the loop, each character (which is a string representation of a digit) is converted back into an integer using `int()`, making it ready for arithmetic operations. Simultaneously, the solution maintains two accumulator variables: `product` initialized to `1` and `sum_digits` initialized to `0`. Initializing `product` to `1` is crucial because multiplying by `0` would incorrectly result in `0` regardless of the other digits (unless `n` itself contained a 0, which would correctly make the product 0). As each digit is extracted and converted to an integer, it's immediately multiplied into `product` and added to `sum_digits`. This single-pass accumulation is highly efficient as it avoids iterating over the digits multiple times. Once the loop finishes processing all digits, both `product` and `sum_digits` hold their respective final values, and the last step is simply returning their difference.

Step-by-Step Algorithm

Step 1: Initialize two variables, `product` and `sum`, to 1 and 0 respectively. `product` will store the product of the digits, and `sum` will store their sum.
Step 2: Iterate through the digits of the input integer `n`. This can be done using a `while` loop (as shown in the Java, C++, and C solutions) or by converting the integer to a string and iterating over its characters (as in the Python solution).
Step 3: In each iteration, extract the last digit using the modulo operator (`n % 10`).
Step 4: Multiply the current `product` by the extracted digit.
Step 5: Add the extracted digit to the current `sum`.
Step 6: Remove the last digit from `n` using integer division (`n /= 10`).
Step 7: Repeat steps 3-6 until `n` becomes 0.
Step 8: After the loop finishes, return the difference between `product` and `sum` (`product - sum`).

Key Insights

Leveraging string conversion (`str(n)`) is an elegant and readable way to iterate through the individual digits of an integer, effectively abstracting away more traditional modulo and division operations which can sometimes be more verbose.
Correctly initializing accumulators is vital: the product accumulator must start at `1` to ensure that initial multiplications don't result in an incorrect zero-out, while the sum accumulator correctly starts at `0`.
The problem can be solved efficiently in a single pass by calculating both the product and sum of digits concurrently within the same loop, rather than performing separate iterations for each aggregation.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Quora.

Subtract the Product and Sum of Digits of an Integer | LeetCode 1281

Easy

Math

Time: O(log(n))Space: O(1)

Solution Code

Available in:

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;
    }
}

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Problem Info

Description

Problem 1281: Subtract the Product and Sum of Digits of an Integer - Detailed explanation and solution approaches.

Topics

Math

Companies

Quora

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1281

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize two variables, `product` and `sum`, to 1 and 0 respectively. `product` will store the product of the digits, and `sum` will store their sum.
2
Step 2: Iterate through the digits of the input integer `n`. This can be done using a `while` loop (as shown in the Java, C++, and C solutions) or by converting the integer to a string and iterating over its characters (as in the Python solution).
3
Step 3: In each iteration, extract the last digit using the modulo operator (`n % 10`).
4
Step 4: Multiply the current `product` by the extracted digit.
5
Step 5: Add the extracted digit to the current `sum`.
6
Step 6: Remove the last digit from `n` using integer division (`n /= 10`).
7
Step 7: Repeat steps 3-6 until `n` becomes 0.
8
Step 8: After the loop finishes, return the difference between `product` and `sum` (`product - sum`).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The number of iterations required is proportional to the number of digits in the input integer `n`. The number of digits is logarithmic with respect to `n` (base 10). The algorithm uses only a constant amount of extra space regardless of the input size; hence, the space complexity is O(1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A recursive function could also be used to solve this problem. The base case would be when n is a single-digit number. The recursive step would extract the last digit, recursively process the remaining number, and then combine the results.

Time: O(log₁₀(n))Space: O(log₁₀(n))

Trade-offs: The recursive approach is less efficient due to function call overhead, but it might be more concise for some programmers. The iterative approach is generally preferred for better performance.

✨

Key Insights

1Leveraging string conversion (`str(n)`) is an elegant and readable way to iterate through the individual digits of an integer, effectively abstracting away more traditional modulo and division operations which can sometimes be more verbose.
2Correctly initializing accumulators is vital: the product accumulator must start at `1` to ensure that initial multiplications don't result in an incorrect zero-out, while the sum accumulator correctly starts at `0`.
3The problem can be solved efficiently in a single pass by calculating both the product and sum of digits concurrently within the same loop, rather than performing separate iterations for each aggregation.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The input `n` is a single-digit number. The solution handles this correctly; the loop will execute only once.
Edge case 2: The input `n` is 100000. The solution correctly processes this largest possible input within the constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the initialization of `product`. It should be initialized to 1, not 0, to ensure correct multiplication.

✗Common mistake 2: Off-by-one errors in the loop condition or digit extraction. Carefully consider the loop termination condition and how to extract each digit.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This type of digit manipulation could be useful in tasks involving checksum verification or other data validation techniques.

🏦Real-world scenario 2: In certain cryptographic algorithms or encoding schemes, manipulation of individual digits within a number might be necessary.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Subtract the Product and Sum of Digits of an Integer"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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