What is the time complexity of Integer Break?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Integer Break?

This problem uses Math, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Integer Break?

Insight 1: The optimal way to break down an integer into a sum to maximize the product is to use as many 3s as possible. Insight 2: The number 2 is preferred to 1 because 2 > 1, but also because 2 * 2 > 1 * 3 > 1 * 1.

Which companies ask LeetCode 343. Integer Break in interviews?

This problem is frequently asked at Accenture.

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Integer Break - LeetCode 343 Solution

Integer Break - Complete Solution Guide

Integer Break is LeetCode problem 343, a Medium 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 n , break it into the sum of k positive integers , where k >= 2 , and maximize the product of those integers. Return the maximum product you can get . Example 1: Input: n = 2 Output: 1 Explanation: 2 = 1 + 1, 1 × 1 = 1. Example 2: Input: n = 10 Output: 36 Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36. Constraints: 2 <= n <= 58

Detailed Explanation

The problem asks us to break down a given integer 'n' into a sum of at least two positive integers (k >= 2) and maximize the product of these integers. For example, if n = 10, we can break it into 3 + 3 + 4, and the product is 3 * 3 * 4 = 36. The goal is to find the best possible breakdown of 'n' that results in the largest product.

Solution Approach

The provided solution utilizes a mathematical approach based on the observation that breaking 'n' into as many 3s as possible maximizes the product. It handles the base cases (n <= 3) directly. For larger 'n', it calculates the remainder when 'n' is divided by 3. Based on the remainder (0, 1, or 2), it calculates the product accordingly. If the remainder is 0, the answer is 3 raised to the power of (n / 3). If the remainder is 1, one '3' is replaced with a '4' (2 * 2). If the remainder is 2, the last part is '3 * 2'.

Step-by-Step Algorithm

Step 1: Handle the base case: If n is less than or equal to 3, return n - 1. This is because 2 = 1 + 1 (1 * 1 = 1) and 3 = 1 + 2 (1 * 2 = 2).
Step 2: Calculate the remainder when n is divided by 3.
Step 3: If the remainder is 0, it means n is divisible by 3. The maximum product is 3^(n/3).
Step 4: If the remainder is 1, it means we have a factor of 1 left after dividing n by 3. Instead of having 3 + 3 + ... + 3 + 1, we convert the last 3 + 1 to 2 + 2. Thus the maximum product is 3^((n/3) - 1) * 4.
Step 5: If the remainder is 2, we have a factor of 2 left. The maximum product is 3^(n/3) * 2.

Key Insights

Insight 1: The optimal way to break down an integer into a sum to maximize the product is to use as many 3s as possible.
Insight 2: The number 2 is preferred to 1 because 2 > 1, but also because 2 * 2 > 1 * 3 > 1 * 1.
Insight 3: When n is small (n <= 3), the answer is n-1. For larger n, we prioritize 3s, but when the remainder is 1 after dividing n by 3, we replace the last '3 + 1' with '2 + 2' because 2*2 > 3*1.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming.

Companies

Asked at: Accenture.

Integer Break | LeetCode 343

Medium

Math Dynamic Programming

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public int integerBreak(int n) {
        if (n <= 3) {
            return n - 1;
        }
        
        int remainder = n % 3;
        if (remainder == 0) {
            return (int) Math.pow(3, n / 3);
        } else if (remainder == 1) {
            return (int) (Math.pow(3, n / 3 - 1) * 4);
        } else { // remainder == 2
            return (int) (Math.pow(3, n / 3) * 2);
        }
    }
}

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Problem Info

Description

Problem 343: Integer Break - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming

Companies

Accenture

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:343

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

5 steps

1
Step 1: Handle the base case: If n is less than or equal to 3, return n - 1. This is because 2 = 1 + 1 (1 * 1 = 1) and 3 = 1 + 2 (1 * 2 = 2).
2
Step 2: Calculate the remainder when n is divided by 3.
3
Step 3: If the remainder is 0, it means n is divisible by 3. The maximum product is 3^(n/3).
4
Step 4: If the remainder is 1, it means we have a factor of 1 left after dividing n by 3. Instead of having 3 + 3 + ... + 3 + 1, we convert the last 3 + 1 to 2 + 2. Thus the maximum product is 3^((n/3) - 1) * 4.
5
Step 5: If the remainder is 2, we have a factor of 2 left. The maximum product is 3^(n/3) * 2.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves a few simple arithmetic operations (modulo, division, and exponentiation). These operations take constant time, regardless of the input size 'n'. Therefore, both the time and space complexities are 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

Dynamic Programming

A dynamic programming approach can also be used to solve this problem. Create a dp array of size n+1. dp[i] stores the maximum product we can get by breaking i. The recurrence relation is dp[i] = max(dp[i], j * max(i - j, dp[i - j])) for all 1 <= j < i.

Time: O(n^2)Space: O(n)

Trade-offs: DP is less efficient (O(n^2)) than the mathematical solution (O(1)). DP is useful for similar problems with more complex constraints where a direct mathematical formula may not be easily derived. The constant time solution leverages the key insight that 3 is optimal.

✨

Key Insights

1Insight 1: The optimal way to break down an integer into a sum to maximize the product is to use as many 3s as possible.
2Insight 2: The number 2 is preferred to 1 because 2 > 1, but also because 2 * 2 > 1 * 3 > 1 * 1.
3Insight 3: When n is small (n <= 3), the answer is n-1. For larger n, we prioritize 3s, but when the remainder is 1 after dividing n by 3, we replace the last '3 + 1' with '2 + 2' because 2*2 > 3*1.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 2. The function returns 2 - 1 = 1, which is correct because 2 = 1 + 1 and 1 * 1 = 1.
Edge case 2: n = 3. The function returns 3 - 1 = 2, which is correct because 3 = 1 + 2 and 1 * 2 = 2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base cases (n <= 3) correctly. These cases require special treatment because simply applying the '3' rule won't yield the correct result.

✗Common mistake 2: Using a loop for calculating powers in languages where a built-in power function is available. Using the built-in power function is usually more efficient.

✗Common mistake 3: Failing to consider that when the remainder is 1, it's better to replace 3 + 1 with 2 + 2.

✅

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: Resource allocation optimization. Suppose you have a fixed amount of resources (e.g., time, budget) to allocate to different tasks, and the product of the task outcomes represents the overall success. Breaking down the resources into optimal amounts for each task to maximize the total 'product' can be modeled using this problem's principle.

🏦Real-world scenario 2: Modeling investment strategies. Dividing a capital investment into smaller parts across various asset classes. Each asset class has a certain return. The goal is to maximize the overall return, which might be proportional to the product of the returns from each allocation of capital.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: November 11, 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

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

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Integer Break"?

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

🤖 Need help with this problem?