Examine the following code snippet which attempts to solve the problem. Identify the line containing the subtle bug that causes incorrect results on some inputs.

medium🐞 Bug Identification Q14 of Q15

Dynamic Programming: Knapsack - Ones and Zeroes (2D Knapsack)

ALine 7: zeros = s.count('0')

BLine 10: for j in range(0, n - ones + 1):

CLine 9: for i in range(0, m - zeros + 1): # iterating forwards

DLine 11: dp[i + zeros][j + ones] = max(dp[i + zeros][j + ones], 1 + dp[i][j])

Step-by-Step Solution

Solution:

Step 1: Understand dp iteration direction
To avoid counting the same string multiple times, dp must be updated backwards (from high to low indices).
Step 2: Identify bug in iteration
Iterating forwards over i and j causes dp states to be reused within the same iteration, leading to overcounting.
Final Answer:
Option C -> Option C
Quick Check:
Backward iteration is required for 0/1 knapsack variants [OK]

Quick Trick: Forward iteration in 0/1 knapsack dp causes double counting [OK]

Common Mistakes:

MISTAKES

Forgetting to iterate dp backwards
Miscounting zeros and ones

Trap Explanation:

PITFALL

Forward iteration looks intuitive but breaks the 0/1 knapsack property of single use per item.

Interviewer Note:

CONTEXT

Tests if candidate knows subtle dp iteration order bugs in knapsack variants.

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More Dynamic Programming: Knapsack Quizzes

Examine the following code snippet which attempts to solve the problem. Identify the line containing the subtle bug that causes incorrect results on some inputs.

Step 1: Understand dp iteration direction

Step 2: Identify bug in iteration

Final Answer:

Quick Check:

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