[Solved] Suppose now you want to count the number of ways to make change but coins can be used at most once each (no reuse). Which modification to the DP approach correctly solves this variant? | Dynamic Programming: Knapsack

Dynamic Programming: Knapsack - Number of Ways to Make Change

Suppose now you want to count the number of ways to make change but coins can be used at most once each (no reuse). Which modification to the DP approach correctly solves this variant?

AUse 1D DP iterating amounts forwards but reset dp array after each coin

BUse 2D DP with dp[i][w] representing ways using first i coins for amount w, iterating amounts forwards

CUse the same 1D DP but iterate amounts backwards from amount down to coin value

DUse greedy approach picking largest coins first until amount is reached

Step-by-Step Solution

Solution:

Step 1: Understand no reuse constraint
Each coin can be used once, so combinations must not count repeated usage of the same coin.
Step 2: Modify DP iteration order
Iterating amounts backwards in 1D DP ensures each coin contributes only once per amount, preventing reuse.
Step 3: Confirm correctness
This approach correctly counts combinations without reuse, unlike forward iteration which allows multiple usage.
Final Answer:
Option C -> Option C
Quick Check:
Backward iteration in 1D DP enforces single usage per coin [OK]

Quick Trick: Backward iteration prevents coin reuse in 1D DP [OK]

Common Mistakes:

MISTAKES

Using forward iteration allows unlimited reuse
Resetting dp array loses accumulated counts
Greedy approach misses many combinations

Trap Explanation:

PITFALL

Forward iteration is intuitive but breaks no-reuse constraint; backward iteration is subtle but correct.

Interviewer Note:

CONTEXT

Tests candidate's ability to adapt DP for usage constraints and iteration order effects.

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More Dynamic Programming: Knapsack Quizzes

Suppose now you want to count the number of ways to make change but coins can be used at most once each (no reuse). Which modification to the DP approach correctly solves this variant?

Step 1: Understand no reuse constraint

Step 2: Modify DP iteration order

Step 3: Confirm correctness

Final Answer:

Quick Check:

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