Coin Change Total Number of Ways in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the ways array with 1 for amount 0.

DSA Python

ways = [0] * (amount + 1)
ways[[1]] = 1

Drag options to blanks, or click blank then click option'

A1

B-1

Camount

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting ways[1] = 1 instead of ways[0] = 1

Initializing all ways to 0 without setting ways[0]

We set ways[0] = 1 because there is exactly one way to make amount 0: use no coins.

2fill in blank

medium

Complete the code to loop through each coin in coins list.

DSA Python

for [1] in coins:
    for x in range(coin, amount + 1):
        ways[x] += ways[x - coin]

Drag options to blanks, or click blank then click option'

Aamount

Bx

Ccoin

Dways

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'x' instead of 'coin' in the outer loop

Using 'amount' as loop variable

We loop through each coin in the coins list using 'for coin in coins'.

3fill in blank

medium

Complete the function to return the number of ways to make the given amount.

DSA Python

return ways[[1]]

Drag options to blanks, or click blank then click option'

Aamount

B1

Camount + 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning ways[0]

Returning amount

ways[amount] contains the total number of combinations to make the target amount.

4fill in blank

hard

Fix the error in the inner loop to correctly update ways array.

DSA Python

for coin in coins:
    for x in range(coin, amount + 1):
        ways[x] [1]= ways[x - coin]

Drag options to blanks, or click blank then click option'

A-

B+

C*

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which overwrites the value

Using '-' which subtracts ways incorrectly

We add ways[x - coin] to ways[x] to accumulate the number of ways to make amount x.

5fill in blank

hard

Set the correct starting index for the inner loop range to ensure proper order.

DSA Python

for coin in coins:
    for x in range([1], amount + 1):
        ways[x] += ways[x - coin]

Drag options to blanks, or click blank then click option'

Acoin

B0

C1

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 0, which can lead to incorrect order or permutations instead of combinations

Starting from 1

Start the inner loop from 'coin' to avoid using the current coin for amounts smaller than itself and to compute combinations correctly.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total number of ways to make a target amount using given coin denominations.

B. The minimum number of coins needed to make a target amount.

C. The maximum value of coins that can be used to reach the target.

D. The total sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible ways to make a target amount using given coins, not minimizing or maximizing coins.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, so answer is total ways to form amount.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing with minimum coin count problem
• Thinking about maximum coin value instead of ways
• Ignoring that order of coins does not matter

2. Which of the following is the correct Python syntax to initialize a list named dp with size 6 filled with zeros?

easy

A. dp = [6] * 0

B. dp = [0, 6]

C. dp = list(0, 6)

D. dp = [0] * 6

Solution

1. Step 1: Understand list initialization

   To create a list of size 6 filled with zeros, multiply a list containing zero by 6.

2. Step 2: Check each option

   dp = [0] * 6 uses correct syntax dp = [0] * 6. Others are invalid or create wrong lists.

3. Final Answer:

   dp = [0] * 6 -> Option D

4. Quick Check:

   List * size creates repeated elements [OK]

Hint: Use [0] * size to create zero-filled list [OK]

Common Mistakes:

• Using commas inside list incorrectly
• Trying to call list() with two arguments
• Multiplying list by zero instead of size

3. What is the output of the following code?

coins = [1, 2]
amount = 3
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
    for x in range(coin, amount + 1):
        dp[x] += dp[x - coin]
print(dp[amount])

medium

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Initialize dp and start loops

   dp = [1,0,0,0] initially. For coin=1, update dp for x=1 to 3 adding dp[x-1].

2. Step 2: Update dp for each coin

   After coin=1, dp = [1,1,1,1]. For coin=2, update dp[x] adding dp[x-2]: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2.

3. Step 3: Final dp array and answer

   dp = [1,1,2,2], so dp[3] = 2 means 2 ways to make amount 3.

4. Final Answer:

   2 -> Option C

5. Quick Check:

   Ways to make 3 with [1,2] = 2 [OK]

Hint: Use dp array to accumulate ways for each coin [OK]

Common Mistakes:

• Mixing order of loops causing wrong counts
• Not initializing dp[0] = 1
• Confusing dp[x] update with assignment instead of addition

4. Find the error in this code snippet for coin change total ways:

coins = [1, 2, 3]
amount = 4
dp = [0] * (amount + 1)
for coin in coins:
    for x in range(coin, amount + 1):
        dp[x] += dp[x - coin]
print(dp[amount])

medium

A. The inner loop range is incorrect

B. dp array is not initialized with dp[0] = 1

C. The outer loop should be over amount, not coins

D. The print statement is outside the loop

Solution

1. Step 1: Check dp initialization

   dp[0] must be 1 to represent one way to make amount zero, but here dp[0] = 0.

2. Step 2: Understand impact of missing dp[0] = 1

   Without dp[0] = 1, dp[x] never updates correctly, so total ways remain zero.

3. Final Answer:

   dp array is not initialized with dp[0] = 1 -> Option B

4. Quick Check:

   dp[0] = 1 is essential [OK]

Hint: Always set dp[0] = 1 before loops [OK]

Common Mistakes:

• Forgetting dp[0] initialization
• Changing loop order incorrectly
• Misunderstanding dp array purpose

5. Given coins = [1, 3, 5] and amount = 7, how many total ways are there to make amount 7? Use the coin change total number of ways approach.

hard

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Initialize dp array and set dp[0] = 1

   dp = [1,0,0,0,0,0,0,0]

2. Step 2: Update dp for coin=1

   For x=1 to 7, dp[x] += dp[x-1], so dp = [1,1,1,1,1,1,1,1]

3. Step 3: Update dp for coin=3

   For x=3 to 7, dp[x] += dp[x-3]: dp[3]=1+1=2, dp[4]=1+1=2, dp[5]=1+1=2, dp[6]=1+2=3, dp[7]=1+2=3

4. Step 4: Update dp for coin=5

   For x=5 to 7, dp[x] += dp[x-5]: dp[5]=2+1=3, dp[6]=3+1=4, dp[7]=3+1=4

5. Final Answer:

   4 -> Option A

6. Quick Check:

   Count ways stepwise with dp array [OK]

Hint: Update dp for each coin sequentially [OK]

Common Mistakes:

• Not updating dp for all coins
• Mixing order of loops causing duplicates
• Forgetting dp[0] = 1 initialization