Coin Change Total Number of Ways in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of the following code that calculates the total number of ways to make change for amount 5 using coins [1, 2, 3]?

DSA Python

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

print(count_ways([1, 2, 3], 5))

A 6

B 5

C 7

D 8

💡 Hint

Think about how many combinations of coins add up to 5 using 1, 2, and 3.

The code uses dynamic programming to count ways. For amount 5 and coins [1,2,3], the total ways are 7.

❓ Predict Output

intermediate

Number of Ways for Amount 4 with Coins [1, 3]

What does this code print for the number of ways to make amount 4 using coins [1, 3]?

DSA Python

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

print(count_ways([1, 3], 4))

A 2

B 4

C 3

D 5

💡 Hint

Try to list all combinations of 1 and 3 coins that sum to 4.

Ways are: 1+1+1+1, 1+3 (order doesn't matter). Total 2 ways.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does this code raise when trying to count ways to make change for amount 5 with coins [1, 2]?

DSA Python

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

print(count_ways([1, 2], 5))

A TypeError: unsupported operand type(s) for +=

B IndexError: list index out of range

C KeyError: 0

D No error, outputs 3

💡 Hint

Check how dp list is initialized and accessed.

dp is initialized with length amount (5), but code accesses dp[5], which is out of range.

❓ Predict Output

advanced

Output of Coin Change Ways with Zero Amount

What is the output of this code when amount is 0 and coins are [1, 2, 3]?

DSA Python

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

print(count_ways([1, 2, 3], 0))

A None

B 0

C Error

D 1

💡 Hint

Think about how many ways to make amount zero.

There is exactly one way to make amount zero: use no coins.

❓ Predict Output

expert

Number of Ways for Large Amount with Multiple Coins

What is the output of this code for amount 10 with coins [2, 5, 3, 6]?

DSA Python

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

print(count_ways([2, 5, 3, 6], 10))

A 5

B 6

C 7

D 8

💡 Hint

Try to count all combinations carefully or trust the dynamic programming approach.

The dp array after processing all coins results in dp[10] = 5 ways.

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