[Solved] Examine the following code snippet for counting ways to make change where each coin can be used unlimited times:def count_ways(coins, amount): dp = [0] * (amount + 1) dp[0] = 1 for coin in coins:... | Dynamic Programming: Knapsack

Dynamic Programming: Knapsack - Number of Ways to Make Change

Examine the following code snippet for counting ways to make change where each coin can be used unlimited times:

def count_ways(coins, amount):
    dp = [0] * (amount + 1)
    dp[0] = 1
    for coin in coins:
        for w in range(amount, coin - 1, -1):
            dp[w] += dp[w - coin]
    return dp[amount]

What is the subtle bug in this implementation?

AThe function should return dp[0] instead of dp[amount]

Bdp[0] should be initialized to 0 instead of 1

CThe coins list should be sorted before processing

DThe inner loop should iterate forwards from coin to amount, not backwards

Step-by-Step Solution

Solution:

Step 1: Understand the DP update order
For unlimited coin usage, the inner loop must iterate forwards to avoid counting duplicates incorrectly.
Step 2: Identify the bug
The code iterates backwards, which is correct only for the case where coins can be used once.
Final Answer:
Option D -> Option D
Quick Check:
Forward iteration ensures combinations are counted correctly [OK]

Quick Trick: Use forward iteration for unlimited coin usage [OK]

Common Mistakes:

MISTAKES

Using backward iteration which is for single-use coins
Incorrect dp[0] initialization
Assuming sorting coins affects correctness

Trap Explanation:

PITFALL

Backward iteration is correct for single-use coins, not unlimited usage

Interviewer Note:

CONTEXT

Tests understanding of iteration order in DP for coin change

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More Dynamic Programming: Knapsack Quizzes

Examine the following code snippet for counting ways to make change where each coin can be used unlimited times:

Step 1: Understand the DP update order

Step 2: Identify the bug

Final Answer:

Quick Check:

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