Coin Change Minimum Coins in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with a large number.

DSA Python

dp = [[1]] * (amount + 1)

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

A0

B1

Cfloat('inf')

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of a large number causes incorrect minimum calculations.

Using -1 can cause logic errors in comparisons.

We use float('inf') to represent an initially unreachable amount.

2fill in blank

medium

Complete the code to set the base case for amount 0.

DSA Python

dp[0] = [1]

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

Afloat('inf')

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to a large number causes wrong results.

Using 1 instead of 0 adds an extra coin unnecessarily.

0 coins are needed to make amount 0.

3fill in blank

hard

Fix the error in the inner loop condition to check if coin can be used.

DSA Python

for coin in coins:
    for x in range(coin, amount + 1):
        if dp[x - coin] != [1]:
            dp[x] = min(dp[x], dp[x - coin] + 1)

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

Afloat('inf')

B0

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 instead of float('inf') causes wrong updates.

Using -1 causes logic errors.

We check if dp[x - coin] is not infinity before using it.

4fill in blank

hard

Fill both blanks to return the result or -1 if no solution.

DSA Python

return dp[amount] if dp[amount] != [1] else [2]

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

Afloat('inf')

B-1

C0

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 for no solution.

Not checking dp[amount] leads to wrong answers.

If dp[amount] is infinity, no solution exists, so return -1.

5fill in blank

hard

Fill all three blanks to complete the coin change function.

DSA Python

def coin_change(coins, amount):
    dp = [[1]] * (amount + 1)
    dp[0] = [2]
    for coin in coins:
        for x in range(coin, amount + 1):
            if dp[x - coin] != [3]:
                dp[x] = min(dp[x], dp[x - coin] + 1)
    return dp[amount] if dp[amount] != float('inf') else -1

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

Afloat('inf')

B0

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Incorrect initialization of dp array.

Wrong base case value.

Not checking dp[x - coin] properly.

Initialize dp with infinity, set dp[0] to 0, and check dp[x - coin] != infinity.

Practice

1. What does the Coin Change Minimum Coins problem ask you to find?
coins = [1, 2, 5], amount = 11

easy

A. The minimum number of coins needed to make the amount

B. The maximum number of coins that can be used

C. The total sum of all coin values

D. The number of different coin types

Solution

1. Step 1: Understand the problem goal

   The problem asks for the fewest coins needed to sum up to the given amount using the available coin denominations.

2. Step 2: Clarify what is asked

   It is not about maximum coins or total sum, but the minimum count of coins to reach the amount.

3. Final Answer:

   The minimum number of coins needed to make the amount -> Option A

4. Quick Check:

   Minimum coins = goal [OK]

Hint: Focus on minimum coins to reach amount, not max or sum [OK]

Common Mistakes:

• Confusing minimum with maximum coins
• Thinking total sum of coins is the answer
• Counting coin types instead of coin count

2. Which of the following is the correct Python syntax to initialize a list dp of size amount + 1 with all values set to amount + 1?

easy

A. dp = [amount + 1] * amount

B. dp = [amount + 1 for i in range(amount)]

C. dp = [amount] * (amount + 1)

D. dp = [amount + 1] * (amount + 1)

Solution

1. Step 1: Understand list initialization

   We want a list of length amount + 1, each element set to amount + 1.

2. Step 2: Check each option

   dp = [amount + 1] * (amount + 1) correctly creates a list of length amount + 1 with all values amount + 1. dp = [amount + 1 for i in range(amount)] creates length amount, not amount + 1. dp = [amount + 1] * amount length is amount, not amount + 1. dp = [amount] * (amount + 1) sets values to amount, not amount + 1.

3. Final Answer:

   dp = [amount + 1] * (amount + 1) -> Option D

4. Quick Check:

   Correct list size and values [OK]

Hint: Use * operator with correct length for list initialization [OK]

Common Mistakes:

• Off-by-one error in list size
• Wrong value assigned to list elements
• Using list comprehension with wrong range

3. What is the output of the following code?

coins = [1, 3, 4]
amount = 6
dp = [7] * (amount + 1)
dp[0] = 0
for i in range(1, amount + 1):
    for coin in coins:
        if coin <= i:
            dp[i] = min(dp[i], dp[i - coin] + 1)
print(dp[amount])

medium

A. 6

B. 2

C. 1

D. 3

Solution

1. Step 1: Trace dp array updates for amount=6

   Initialize dp = [0,7,7,7,7,7,7]. For i=1: dp[1]=min(7, dp[0]+1)=1. For i=2: dp[2]=min(7, dp[1]+1)=2. For i=3: dp[3]=min(7, dp[0]+1)=1. For i=4: dp[4]=min(7, dp[0]+1)=1. For i=5: dp[5]=min(7, dp[4]+1)=2. For i=6: dp[6]=min(7, dp[5]+1=3, dp[3]+1=2, dp[2]+1=3) so dp[6]=2.

2. Step 2: Final dp[6] value

   The minimum coins to make 6 is 2 (e.g., 3+3 or 4+1+1 but 2 coins is minimal).

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array updates step-by-step for each amount [OK]

Common Mistakes:

• Not updating dp[i] correctly inside nested loops
• Confusing coin values with indices
• Forgetting to initialize dp[0] = 0

4. Find the error in the following code snippet for Coin Change Minimum Coins:

coins = [1, 2, 5]
amount = 11
dp = [amount + 1] * amount
dp[0] = 0
for i in range(1, amount + 1):
    for coin in coins:
        if coin <= i:
            dp[i] = min(dp[i], dp[i - coin] + 1)
print(dp[amount])

medium

A. The inner loop should start from coin = 0

B. dp[0] should be set to amount + 1

C. dp list size is incorrect, should be amount + 1

D. The print statement should print dp[amount - 1]

Solution

1. Step 1: Check dp list initialization size

   dp is initialized with size amount, but indices go up to amount (11), so dp[11] access causes error.

2. Step 2: Correct dp size and consequences

   dp should be size amount + 1 to include index amount. dp[0] = 0 is correct. Other parts are fine.

3. Final Answer:

   dp list size is incorrect, should be amount + 1 -> Option C

4. Quick Check:

   dp size = amount + 1 [OK]

Hint: dp size must cover all indices up to amount [OK]

Common Mistakes:

• Off-by-one error in dp list size
• Incorrect initialization of dp[0]
• Wrong index in print statement

5. Given coins = [2, 5] and amount = 7, which of the following code snippets correctly returns the minimum number of coins needed or -1 if not possible?

hard

A. def min_coins(coins, amount):
dp = [amount + 1] * (amount + 1)
dp[0] = 0
for i in range(1, amount + 1):
for coin in coins:
if coin <= i:
dp[i] = min(dp[i], dp[i - coin] + 1)
return dp[amount] if dp[amount] != amount + 1 else -1

B. def min_coins(coins, amount):
dp = [0] * (amount + 1)
for i in range(1, amount + 1):
dp[i] = min(dp[i - coin] + 1 for coin in coins if coin <= i)
return dp[amount]

C. def min_coins(coins, amount):
dp = [float('inf')] * amount
dp[0] = 0
for i in range(1, amount):
for coin in coins:
if coin <= i:
dp[i] = min(dp[i], dp[i - coin] + 1)
return dp[amount - 1]

D. def min_coins(coins, amount):
dp = [amount] * (amount + 1)
dp[0] = 1
for i in range(1, amount + 1):
for coin in coins:
if coin < i:
dp[i] = min(dp[i], dp[i - coin] + 1)
return dp[amount]

Solution

1. Step 1: Analyze each code snippet

   def min_coins(coins, amount): dp = [amount + 1] * (amount + 1) dp[0] = 0 for i in range(1, amount + 1): for coin in coins: if coin <= i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount] if dp[amount] != amount + 1 else -1 correctly initializes dp with amount + 1 size and value, sets dp[0]=0, updates dp[i] properly, and returns -1 if no solution. def min_coins(coins, amount): dp = [0] * (amount + 1) for i in range(1, amount + 1): dp[i] = min(dp[i - coin] + 1 for coin in coins if coin <= i) return dp[amount] lacks initialization for dp[0] and may cause error if no valid coins. def min_coins(coins, amount): dp = [float('inf')] * amount dp[0] = 0 for i in range(1, amount): for coin in coins: if coin <= i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount - 1] uses wrong dp size and indices, and returns dp[amount - 1] incorrectly. def min_coins(coins, amount): dp = [amount] * (amount + 1) dp[0] = 1 for i in range(1, amount + 1): for coin in coins: if coin < i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount] sets dp[0]=1 incorrectly and uses wrong coin comparison (coin < i instead of coin <= i).

2. Step 2: Confirm correctness of def min_coins(coins, amount): dp = [amount + 1] * (amount + 1) dp[0] = 0 for i in range(1, amount + 1): for coin in coins: if coin <= i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount] if dp[amount] != amount + 1 else -1

   def min_coins(coins, amount): dp = [amount + 1] * (amount + 1) dp[0] = 0 for i in range(1, amount + 1): for coin in coins: if coin <= i: dp[i] = min(dp[i], dp[i - coin] + 1) return dp[amount] if dp[amount] != amount + 1 else -1 matches the standard dynamic programming solution for minimum coins with proper edge case handling.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Standard DP solution with correct initialization and return [OK]

Hint: Check dp size, dp[0], coin comparison, and return condition [OK]

Common Mistakes:

• Wrong dp list size or initialization
• Incorrect dp[0] value
• Using coin < i instead of coin <= i
• Not handling no solution case (-1)