#343

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Integer Break

Break an integer into the sum of positive integers to maximize the product using dynamic programming.

Problem Statement

Given an integer n, break it into the sum of k positive integers where k >= 2, such that the product of these integers is maximized. Your task is to return the maximum product that can be obtained by breaking the number n into at least two positive integers.

For example, when n = 2, the only option is 1 + 1, yielding a product of 1. Similarly, for n = 10, the optimal breakdown is 3 + 3 + 4, which results in the maximum product of 36.

Examples

Example 1

Input: n = 2

Output: 1

2 = 1 + 1, 1 × 1 = 1.

Example 2

Input: n = 10

Output: 36

10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Constraints

2 <= n <= 58

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to store the maximum product for each integer from 1 to n. For each i, consider the possible ways to split i into two parts, and track the maximum product. Transition between states by updating the product for smaller subproblems, optimizing the solution incrementally.

Greedy Approach for Optimal Product

Mathematically, the best strategy is to break n into as many 3s as possible, and handle the remainder with 2s. This is because 3 is the most efficient number to maximize the product due to its properties in multiplication. For example, 10 = 3 + 3 + 4 is optimal, as 3 × 3 × 4 = 36.

Memoization to Improve Efficiency

Memoization can be used to avoid redundant calculations by storing already computed results for integer breakdowns. This reduces the complexity by eliminating repeated work, enabling a faster solution even for larger values of n.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n) with the dynamic programming approach, as we compute the maximum product for each number up to n once. The space complexity is also O(n) due to the storage of intermediate results.

What Interviewers Usually Probe

Does the candidate recognize the optimal use of 3s to maximize the product?
Can they effectively use dynamic programming or memoization to solve the problem efficiently?
Are they able to optimize the solution for larger values of n without redundant calculations?

Common Pitfalls or Variants

Common pitfalls

Forgetting that the product is maximized by breaking n into 3s and 2s, not just splitting n evenly.
Overcomplicating the problem by attempting brute-force enumeration instead of dynamic programming.
Not considering edge cases like n = 2 or n = 3 where the breakdown may seem trivial but is important.

Follow-up variants

Generalizing to finding the maximum product for other mathematical break-downs, such as sums of squares.
Changing the problem constraints to allow for fractional parts, altering the breakdown strategy.
Adapting this problem to work with larger constraints like n = 100 and ensuring that the solution still works efficiently.

FAQ

What is the best way to break an integer to maximize the product?

The most efficient way is to break the integer into as many 3s as possible and handle any remainder with 2s.

How does dynamic programming help in solving the Integer Break problem?

Dynamic programming stores intermediate results to avoid redundant calculations, allowing for an efficient solution to the problem.

What is the time complexity of solving the Integer Break problem using dynamic programming?

The time complexity is O(n) due to the need to compute the maximum product for each number up to n only once.

Can this problem be solved in less than `O(n)` time?

No, the solution involves processing each value up to n to find the optimal breakdown, so O(n) is the best achievable time complexity.

What is the space complexity of the Integer Break dynamic programming approach?

The space complexity is O(n) as it requires storage for intermediate results for each integer from 1 to n.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i]$ as the maximum product that can be obtained by splitting the positive integer $i$, with an initial condition of $f[1] = 1$. The answer is $f[n]$.

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class Solution:
    def integerBreak(self, n: int) -> int:
        f = [1] * (n + 1)
        for i in range(2, n + 1):
            for j in range(1, i):
                f[i] = max(f[i], f[i - j] * j, (i - j) * j)
        return f[n]

Solution 1: Mathematics

When $n < 4$, since the problem requires splitting into at least two integers, $n - 1$ yields the maximum product. When $n \geq 4$, we split into as many $3$s as possible. If the last segment remaining is $4$, we split it into $2 + 2$ for the maximum product.

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class Solution:
    def integerBreak(self, n: int) -> int:
        f = [1] * (n + 1)
        for i in range(2, n + 1):
            for j in range(1, i):
                f[i] = max(f[i], f[i - j] * j, (i - j) * j)
        return f[n]

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