Suppose the coin change problem is modified so that each coin can only be used at most once (0/1 knapsack variant). Which of the following changes to the bottom-up DP approach correctly solves this variant?
AUse a 2D dp array where dp[i][j] represents minimum coins to make amount j using first i coins, iterating coins outer and amounts inner
BSort coins and greedily pick largest coins without repetition until amount is reached
CKeep the same 1D dp array and iterate coins inside the amount loop as before (unbounded usage)
DUse the same 1D dp but iterate amounts outer and coins inner, updating dp[j] only if dp[j - coin] was from previous iteration
Step-by-Step Solution
Solution:
Step 1: Understand the 0/1 usage constraint
Each coin can be used once, so we must track usage per coin, requiring 2D dp.
Step 2: Identify correct DP structure
2D dp with dp[i][j] = min coins to make j using first i coins ensures no reuse; iterate coins outer, amounts inner.
Step 3: Why other options fail
Keep the same 1D dp array and iterate coins inside the amount loop as before (unbounded usage) allows unlimited reuse; B is greedy and incorrect; D misuses 1D dp which is for unbounded usage.
Final Answer:
Option A -> Option A
Quick Check:
0/1 knapsack requires 2D dp to track coin usage [OK]
Quick Trick:0/1 knapsack needs 2D dp to prevent reuse [OK]
Common Mistakes:
MISTAKES
Using unbounded knapsack DP for 0/1 problem
Assuming greedy works for minimum coins
Trap Explanation:
PITFALL
Candidates often reuse unbounded DP code without adjusting for single usage constraint, leading to wrong answers.
Interviewer Note:
CONTEXT
Tests candidate's ability to adapt DP approach to problem variants and constraints.
Master "Coin Change (Minimum Coins)" in Dynamic Programming: Knapsack
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