[Solved] Suppose the integer break problem is modified so that you can reuse the same integer parts multiple times (unbounded... Which modification to the DP approach below correctly handles this variant? | Dynamic Programming: Knapsack

Dynamic Programming: Knapsack - Integer Break

Suppose the integer break problem is modified so that you can reuse the same integer parts multiple times (unbounded splits), and you want to find the maximum product. Which modification to the DP approach below correctly handles this variant?

def integer_break(n):
    dp = [0] * (n + 1)
    dp[1] = 1
    for i in range(2, n + 1):
        max_product = 0
        for j in range(1, i):
            max_product = max(max_product, max(j, dp[j]) * max(i - j, dp[i - j]))
        dp[i] = max_product
    return dp[n]

Options:

AUse a nested loop where for each i, iterate j from 1 to i and update dp[i] = max(dp[i], j * dp[i-j]) to allow reuse

BSort possible parts and greedily pick the largest part repeatedly to maximize product

CKeep the same code but add memoization to avoid recomputation

DChange inner loop to iterate over all j ≤ i and update dp[i] as max(dp[i], dp[i-j] * dp[j]) without max(j, dp[j])

Step-by-Step Solution

Step 1: Understand reuse requirement
Allowing reuse means dp[i] depends on dp[i-j] multiplied by j (or dp[j]) where j can be reused multiple times.
Step 2: Modify DP update
Updating dp[i] = max(dp[i], j * dp[i-j]) correctly models unbounded splits by combining current part j and best product for remaining i-j.
Step 3: Evaluate other options
Change inner loop to iterate over all j ≤ i and update dp[i] as max(dp[i], dp[i-j] * dp[j]) without max(j, dp[j]) removes max(j, dp[j]) incorrectly; Keep the same code but add memoization to avoid recomputation adds memoization but doesn't change logic; Sort possible parts and greedily pick the largest part repeatedly to maximize product is greedy and fails on some inputs.
Final Answer:
Option A -> Option A
Quick Check:
Unbounded knapsack style DP uses dp[i] = max(dp[i], j * dp[i-j]) [OK]

Quick Trick: Unbounded knapsack uses dp[i] = max(dp[i], j * dp[i-j]) [OK]

Common Mistakes:

MISTAKES

Using greedy approach
Not adjusting DP recurrence for reuse

Trap Explanation:

PITFALL

Naive DP or greedy fails to handle reuse; correct recurrence must reflect unbounded splits.

Interviewer Note:

CONTEXT

Tests if candidate can adapt DP to variants requiring reuse of parts

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More Dynamic Programming: Knapsack Quizzes

Which modification to the DP approach below correctly handles this variant?

Step 1: Understand reuse requirement

Step 2: Modify DP update

Step 3: Evaluate other options

Final Answer:

Quick Check:

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