[Solved] Suppose the problem changes so that you can use each item an unlimited number of times (unbounded knapsack). Which modification to the space-optimized bottom-up DP code correctly solves this variant? | Dynamic Programming: Knapsack

Dynamic Programming: Knapsack - 0/1 Knapsack Problem

Suppose the problem changes so that you can use each item an unlimited number of times (unbounded knapsack). Which modification to the space-optimized bottom-up DP code correctly solves this variant?

AChange the inner loop to iterate forwards from weights[i] up to W.

BAdd memoization to the recursive solution to avoid recomputation.

CChange the inner loop to iterate backwards from W down to weights[i].

DUse a greedy approach selecting items with the highest value-to-weight ratio repeatedly.

Step-by-Step Solution

Solution:

Step 1: Understand difference between 0/1 and unbounded knapsack
In unbounded knapsack, items can be chosen multiple times, so dp[w] can be updated using dp[w - weights[i]] from the same iteration.
Step 2: Identify correct iteration order
Iterating forwards from weights[i] to W allows reuse of updated dp values within the same iteration, enabling multiple counts of the same item.
Step 3: Why other options fail
Backward iteration prevents multiple usage in one iteration (wrong for unbounded). Memoization helps but is not the direct fix. Greedy fails for 0/1 and unbounded knapsack except fractional case.
Final Answer:
Option A -> Option A
Quick Check:
Forward iteration enables unlimited item reuse [OK]

Quick Trick: Unbounded knapsack requires forward iteration in dp updates [OK]

Common Mistakes:

MISTAKES

Using backward iteration from 0/1 knapsack causes incorrect results.

Trap Explanation:

PITFALL

Candidates often reuse 0/1 knapsack code without changing iteration direction, breaking unbounded logic.

Interviewer Note:

CONTEXT

Tests candidate's ability to adapt DP approach to problem variants and understand iteration impact.

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More Dynamic Programming: Knapsack Quizzes

Suppose the problem changes so that you can use each item an unlimited number of times (unbounded knapsack). Which modification to the space-optimized bottom-up DP code correctly solves this variant?

Step 1: Understand difference between 0/1 and unbounded knapsack

Step 2: Identify correct iteration order

Step 3: Why other options fail

Final Answer:

Quick Check:

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