Coin Change Minimum Coins in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimum coins for amount 11

What is the output of the following code that calculates the minimum number of coins needed to make amount 11 using coins [1, 2, 5]?

DSA Python

def min_coins(coins, amount):
    dp = [float('inf')] * (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] != float('inf') else -1

print(min_coins([1, 2, 5], 11))

A -1

B 4

C 5

D 3

💡 Hint

Think about how many coins of 5, 2, and 1 can sum to 11 with minimum count.

The minimum coins to make 11 are 5 + 5 + 1, which is 3 coins.

❓ Predict Output

intermediate

Minimum coins for amount 7 with coins [2, 4]

What is the output of the code below for amount 7 using coins [2, 4]?

DSA Python

def min_coins(coins, amount):
    dp = [float('inf')] * (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] != float('inf') else -1

print(min_coins([2, 4], 7))

A -1

B 3

C 2

D 1

💡 Hint

Check if 7 can be formed by any combination of 2 and 4.

7 cannot be formed by any combination of 2 and 4, so the output is -1.

🔧 Debug

advanced

Identify the error in coin change code

What error does the following code produce when run?

DSA Python

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

print(min_coins([1, 2, 5], 11))

A No error, output is 3

B IndexError

C TypeError

D ValueError

💡 Hint

Check initialization of dp array and how min function is used.

The code runs correctly and outputs 3, the minimum coins for 11.

🚀 Application

advanced

Minimum coins for amount 15 with coins [1, 3, 7]

What is the minimum number of coins needed to make amount 15 using coins [1, 3, 7]?

DSA Python

def min_coins(coins, amount):
    dp = [float('inf')] * (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] != float('inf') else -1

print(min_coins([1, 3, 7], 15))

A 4

B 3

C 5

D 6

💡 Hint

Try to combine 7, 3, and 1 coins to reach 15 with fewest coins.

The minimum coins to make 15 are 7 + 7 + 1, which is 3 coins.

🧠 Conceptual

expert

Why does greedy fail for coin change minimum coins?

Why does a greedy approach (always picking the largest coin first) sometimes fail to find the minimum number of coins needed?

A Because greedy always finds the minimum coins for any coin set

B Because greedy approach is slower than dynamic programming

C Because the largest coin may not fit evenly and smaller coins can combine to fewer total coins

D Because greedy uses recursion which causes stack overflow

💡 Hint

Think about cases where picking the largest coin first leads to more coins overall.

Greedy fails when picking the largest coin first leads to leftover amounts that require many small coins, while a combination of smaller coins could use fewer coins overall.

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)