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Suppose the stones can be used multiple times (unlimited quantity of each stone). Which modification to the DP approach correctly computes the minimal last stone weight?

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Dynamic Programming: Knapsack - Last Stone Weight II
Suppose the stones can be used multiple times (unlimited quantity of each stone). Which modification to the DP approach correctly computes the minimal last stone weight?
AUse a greedy approach smashing largest stones repeatedly
BKeep the inner loop iterating backwards as in 0/1 knapsack to avoid reuse
CChange the inner loop to iterate forwards from weight to half, allowing reuse of stones
DUse recursion without memoization to explore all combinations with reuse
Step-by-Step Solution

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

    For unlimited reuse, inner loop must iterate forwards to allow multiple counts of the same stone.

  2. Step 2: Modify DP accordingly

    Iterating forwards from weight to half updates dp[w] using dp[w - weight] from the same iteration, enabling reuse.

  3. Step 3: Why other options fail

    Backward iteration prevents reuse; greedy is incorrect; recursion without memoization is inefficient.

  4. Final Answer:

    Option C -> Option C

  5. Quick Check:

    Forward iteration in inner loop enables unlimited stone reuse [OK]
Quick Trick: Unbounded knapsack -> inner loop forwards [OK]
Common Mistakes:
MISTAKES
  • Using backward iteration causing no reuse
  • Trying greedy which is incorrect
  • Ignoring memoization leading to TLE
Trap Explanation:
PITFALL
  • Backward iteration is standard for 0/1 knapsack; forward iteration is subtle but required for unbounded knapsack.
Interviewer Note:
CONTEXT
  • Tests if candidate understands knapsack variants and how iteration order affects reuse.
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