LeetCode 343: Integer Break Solution in Python – A Step-by-Step Guide

What Is LeetCode 343: Integer Break?

In LeetCode 343: Integer Break, you’re given an integer n, and your task is to break it into at least two positive integers whose sum is n, maximizing their product. For example, with n = 10, the maximum product is 36 (3+3+4 = 10, 334 = 36). This problem tests your ability to optimize integer partitions, connecting to concepts in LeetCode 279: Perfect Squares.

Problem Statement

• Input: An integer n.
• Output: An integer—the maximum product achievable by breaking n into at least two parts.
• Rules:
• Break n into at least two positive integers.
• Sum of parts equals n.
• Maximize the product of the parts.

Constraints

• 2 <= n <= 58

Examples

• Input: 2
• Output: 1 (2 = 1+1, 1*1 = 1)
• Input: 10
• Output: 36 (10 = 3+3+4, 3*3*4 = 36)
• Input: 8
• Output: 18 (8 = 2+3+3, 2*3*3 = 18)

Understanding the Problem: Maximizing the Product

To solve LeetCode 343: Integer Break in Python, we need to split n into at least two positive integers and maximize their product. A naive approach—trying all partitions—is O(2^n), exponential and too slow for n up to 58. Instead, we’ll use:

• Best Solution (Mathematical): O(1) time, O(1) space—fast and insightful.
• Alternative Solution (Dynamic Programming): O(n²) time, O(n) space—clear but less efficient.

Let’s start with the mathematical solution, explained in a beginner-friendly way.

Best Solution: Mathematical Approach

Why This Is the Best Solution

The mathematical approach is the top choice for LeetCode 343 because it’s blazing fast—O(1) time, O(1) space—and leverages a key insight: breaking numbers into 3s (and sometimes 2s) maximizes the product, derived from mathematical properties. It avoids iteration by using a simple formula based on powers of 3. It’s like a master splitter with a secret formula—smart and quick!

How It Works

Think of this as a number splitter:

• Step 1: Key Insight:
• For any n > 4, breaking into 3s maximizes product (e.g., 6 = 3+3 > 2+2+2).
• Use 2s only for small remainders (2 or 4).
• Step 2: Compute Parts:
• Divide n by 3: quotient (q) and remainder (r).
• Cases:
• r = 0: Product = 3^q.
• r = 1: Adjust to 3^(q-1) * 4 (since 3+1 = 4 > 3*1).
• r = 2: Product = 3^q * 2.
• Step 3: Handle Small n:
• n = 2: 1*1 = 1.
• n = 3: 1*2 = 2.
• Step 4: Why It Works:
• 3 is optimal (maximizes x^(1/x), peaks near 3).
• Formula covers all cases efficiently.

It’s like a math-powered splitting machine!

Step-by-Step Example

Example: n = 10

• Divide: 10 // 3 = 3 (q), 10 % 3 = 1 (r)
• Case r=1: q-1 = 2, Product = 3² * 4 = 9 * 4 = 36
• Check: 3+3+4 = 10, 3*3*4 = 36
• Result: 36

Example: n = 8

• Divide: 8 // 3 = 2 (q), 8 % 3 = 2 (r)
• Case r=2: Product = 3² * 2 = 9 * 2 = 18
• Check: 3+3+2 = 8, 3*3*2 = 18
• Result: 18

Code with Detailed Line-by-Line Explanation

Copy codeclass Solution:
    def integerBreak(self, n: int) -> int:
        # Handle small cases
        if n == 2:
            return 1  # 1*1
        if n == 3:
            return 2  # 1*2

        # Compute quotient and remainder
        q, r = divmod(n, 3)

        # Apply formula based on remainder
        if r == 0:
            return 3 ** q  # All 3s
        elif r == 1:
            return 3 ** (q - 1) * 4  # Replace last 3+1 with 4
        else:  # r == 2
            return 3 ** q * 2  # Add a 2

• Lines 3-6: Handle n=2 and n=3 directly.
• Line 9: Compute quotient (q) and remainder (r) of n/3.
• Lines 12-16: Apply formula:
• r=0: Pure 3s.
• r=1: Adjust one 3 to 4.
• r=2: Append a 2.
• Time Complexity: O(1)—constant-time math operations.
• Space Complexity: O(1)—no extra space.

This is like a math-powered product maximizer—instant and sleek!

Alternative Solution: Dynamic Programming

Why an Alternative Approach?

The dynamic programming approach—O(n²) time, O(n) space—builds the maximum product for each number up to n by trying all possible splits. It’s more intuitive for understanding the recursive nature of the problem but slower and space-heavy. It’s like testing every split step-by-step—thorough but bulky!

How It Works

Build up with DP:

• Step 1: dp[i] = max product for breaking i.
• Step 2: For each i:
• Try all splits j + (i-j), max of j * (i-j) and j * dp[i-j].
• Step 3: Return dp[n].

Step-by-Step Example

Example: n = 4

• Init: dp = [0, 0, 1, 2, 0]
• i=4:
• 1+3: 1 * dp[3] = 1*2 = 2
• 2+2: 2 * dp[2] = 2*1 = 2
• 3+1: 3 * dp[1] = 3*0 = 0
• dp[4] = max(2, 2, 0, 2*2) = 4
• Result: 4 (2*2)

Code for DP Approach

Copy codeclass Solution:
    def integerBreak(self, n: int) -> int:
        dp = [0] * (n + 1)
        dp[2] = 1  # Base case

        for i in range(3, n + 1):
            for j in range(1, i):
                dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))

        return dp[n]

• Time Complexity: O(n²)—nested loops.
• Space Complexity: O(n)—dp array.

It’s a step-by-step product builder—vivid but heavy!

Comparing the Two Solutions

• Mathematical (Best):
• Pros: O(1) time, O(1) space, instant solution.
• Cons: Requires math insight.
• Dynamic Programming (Alternative):
• Pros: O(n²) time, O(n) space, intuitive buildup.
• Cons: Slower and space-heavy.

Mathematical wins for efficiency.

Additional Examples and Edge Cases

• 3: 2.
• 5: 6 (2*3).
• 2: 1.

Both handle these, but mathematical is faster.

Complexity Breakdown

• Mathematical: Time O(1), Space O(1).
• DP: Time O(n²), Space O(n).

Mathematical is the clear choice.

Key Takeaways

• Mathematical: Formula magic—fast!
• DP: Build step-by-step—clear!
• Python: Math and arrays shine—see [Python Basics](/python/basics).
• Breaking: 3s rule the game.

Final Thoughts: Break for the Max

LeetCode 343: Integer Break in Python is a mathematical optimization delight. The mathematical solution offers instant brilliance, while DP provides a tangible process. Want more number challenges? Try LeetCode 279: Perfect Squares or LeetCode 343: Integer Break. Ready to split? Head to Solve LeetCode 343 on LeetCode and maximize that product today!