[Solved] In the following bottom-up DP code for integer break, which line causes incorrect results... In the following bottom-up DP code for integer break, which line causes incorrect results when n = 2? | Dynamic Programming: Knapsack

Dynamic Programming: Knapsack - Integer Break

In the following bottom-up DP code for integer break, which line causes incorrect results when n = 2?

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

Afor j in range(1, i):

Bfor i in range(2, n + 1):

Cdp[1] = 1

Ddp[i] = max(dp[i], max(j, dp[j]) * (i - j))

Step-by-Step Solution

Solution:

Step 1: Understand dp initialization
Setting dp[1] = 1 is incorrect because integer break requires at least two parts.
Step 2: Impact on n=2
For n=2, dp[2] depends on dp[1], which should be 0 to avoid invalid splits.
Final Answer:
Option C -> Option C
Quick Check:
dp[1] must be 0 to ensure valid splits [OK]

Quick Trick: dp[1] should be zero for integer break base case [OK]

Common Mistakes:

MISTAKES

Initializing dp[1] to 1 instead of 0
Misunderstanding base cases in DP
Assuming dp[1] represents a valid split product

Trap Explanation:

PITFALL

dp[1] = 1 looks intuitive but breaks correctness for small n

Interviewer Note:

CONTEXT

Tests understanding of DP base cases and correctness

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More Dynamic Programming: Knapsack Quizzes

In the following bottom-up DP code for integer break, which line causes incorrect results when n = 2?

Step 1: Understand dp initialization

Step 2: Impact on n=2

Final Answer:

Quick Check:

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