Suppose the problem is modified so that each coin can be used at most once (0/1 knapsack variant). Which of the following changes to the original code correctly counts the number of combinations?

hard🎤 Interviewer Follow-up Q15 of Q15

Dynamic Programming: Knapsack - Coin Change II (Count Ways)

AChange inner loop to iterate amounts backward (from amount down to coin) to avoid reuse

BChange inner loop to iterate amounts forward (from coin to amount) as in original code

CUse recursion without memoization to explore all subsets

DUse greedy approach picking largest coins first

Step-by-Step Solution

Solution:

Step 1: Understand 0/1 knapsack constraints
Each coin can be used once, so dp updates must avoid reuse within same iteration.
Step 2: Identify correct iteration order
Iterating amounts backward ensures dp[w] uses dp[w - coin] from previous iteration, preventing reuse.
Step 3: Evaluate other options
Forward iteration allows reuse; recursion without memoization is inefficient; greedy is incorrect.
Final Answer:
Option A -> Option A
Quick Check:
Backward iteration is standard for 0/1 knapsack to avoid reuse [OK]

Quick Trick: Backward iteration prevents reuse in 0/1 knapsack [OK]

Common Mistakes:

MISTAKES

Using forward iteration causing reuse
Relying on greedy or naive recursion

Trap Explanation:

PITFALL

Forward iteration looks natural but breaks 0/1 constraint by allowing reuse.

Interviewer Note:

CONTEXT

Tests candidate's understanding of iteration order impact on reuse constraints.

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More Dynamic Programming: Knapsack Quizzes

Suppose the problem is modified so that each coin can be used at most once (0/1 knapsack variant). Which of the following changes to the original code correctly counts the number of combinations?

Step 1: Understand 0/1 knapsack constraints

Step 2: Identify correct iteration order

Step 3: Evaluate other options

Final Answer:

Quick Check:

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