Integer Break - Practice Coding Problems

Integer Break - Problem

You're given a positive integer n, and your task is to break it into the sum of at least 2 positive integers such that the product of those integers is maximized.

For example, if n = 8, you could break it as:

8 = 1 + 7 → product = 1 × 7 = 7
8 = 2 + 6 → product = 2 × 6 = 12
8 = 3 + 5 → product = 3 × 5 = 15
8 = 2 + 3 + 3 → product = 2 × 3 × 3 = 18 ✓ (maximum!)

Your goal is to find the maximum possible product you can achieve by breaking the integer optimally.

Input & Output

example_1.py — Basic Case

$ Input: n = 2

› Output: 1

💡 Note: 2 = 1 + 1, and the product is 1 × 1 = 1

example_2.py — Optimal Split

$ Input: n = 10

› Output: 36

💡 Note: 10 = 3 + 3 + 2 + 2, and the product is 3 × 3 × 2 × 2 = 36. Alternative: 10 = 4 + 3 + 3 gives 4 × 3 × 3 = 36

example_3.py — Edge Case

$ Input: n = 4

› Output: 4

💡 Note: 4 = 2 + 2, and the product is 2 × 2 = 4. Note that 4 = 1 + 3 gives 1 × 3 = 3, which is smaller

Constraints

2 ≤ n ≤ 58
You must break n into at least 2 positive integers
All integers in the sum must be positive

Visualization

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Understanding the Visualization

Try different splits

For n=8, try: 1+7=7, 2+6=12, 3+5=15, 4+4=16

Consider more splits

Try: 2+2+4=16, 2+3+3=18, 3+3+2=18

Find the pattern

Notice that using more 3s tends to give better results

Apply mathematical insight

3 is the magic number - use as many as possible!

Key Takeaway

🎯 Key Insight: The number 3 provides the optimal balance - it maximizes the product-to-sum ratio, making it the ideal building block for achieving maximum product.

Asked in

a Amazon 15 G Google 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses a mathematical insight: to maximize the product, use as many 3s as possible. This is because 3 minimizes the ratio number/ln(number), leading to maximum efficiency. Handle remainders by using 2s when needed, achieving O(1) time and space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Pattern (Optimal)	O(1)	O(1)	Use mathematical insight that 3s give the maximum product per unit sum
Brute Force (Recursive)	O(2^n)	O(n)	Try all possible ways to break the integer recursively
Dynamic Programming (Memoization)	O(n²)	O(n)	Use memoization to avoid recalculating the same subproblems

Mathematical Pattern (Optimal) — Algorithm Steps

Handle base cases: n=2 returns 1, n=3 returns 2, n=4 returns 4
For n > 4, calculate how many 3s we can use: quotient = n // 3
Check the remainder: if remainder is 0, use all 3s
If remainder is 1, use one less 3 and add two 2s (since 2×2 > 3×1)
If remainder is 2, use all 3s and add one 2

Visualization

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Step-by-Step Walkthrough

Compare ratios

Compare value/log(value) ratios for different numbers

Identify pattern

Notice that 3 gives the best ratio

Handle remainders

Deal with cases where n is not divisible by 3

Apply formula

Use the mathematical formula for O(1) solution

Code -

solution.c — C

#include <stdio.h>
#include <math.h>

int integerBreak(int n) {
    if (n == 2) return 1;
    if (n == 3) return 2;
    if (n == 4) return 4;
    
    int quotient = n / 3;
    int remainder = n % 3;
    
    if (remainder == 0) {
        return (int)pow(3, quotient);
    } else if (remainder == 1) {
        return (int)pow(3, quotient - 1) * 4;
    } else {
        return (int)pow(3, quotient) * 2;
    }
}

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

Time & Space Complexity

Time Complexity

⏱️

O(1)

Just a few mathematical operations and one power calculation

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables, no additional data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Pattern (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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