Coin Change Minimum Coins in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the goal of the Coin Change Minimum Coins problem?

To find the smallest number of coins needed to make a given amount using available coin denominations.

beginner

What does a dynamic programming approach use to solve the Coin Change Minimum Coins problem?

It uses a table (array) to store the minimum coins needed for all amounts from 0 up to the target amount, building up the solution step-by-step.

intermediate

Why do we initialize the DP array with a large number (like amount+1) in the Coin Change problem?

Because it represents an impossible high number of coins, so any real solution will be smaller and can replace it during updates.

beginner

In the Coin Change Minimum Coins problem, what does dp[0] = 0 mean?

It means zero coins are needed to make the amount 0, which is the base case for building the solution.

intermediate

What is the time complexity of the dynamic programming solution for the Coin Change Minimum Coins problem?

O(n * amount), where n is the number of coin denominations and amount is the target amount.

What does dp[i] represent in the Coin Change Minimum Coins dynamic programming solution?

A The minimum number of coins needed to make amount i

B The maximum number of coins needed to make amount i

C The total number of coins available

D The coin denomination at index i

dp[i] stores the smallest number of coins needed to make the amount i.

What should you do if a coin denomination is larger than the current amount i in the DP loop?

A Skip that coin and continue

B Use that coin anyway

C Reset dp[i] to zero

D Stop the entire algorithm

You skip coins larger than the current amount because they cannot be used to make that amount.

If the dp array value for the target amount remains the initial large number, what does it mean?

A The amount is zero

B It is impossible to make the amount with given coins

C The coins are unlimited

D The amount is negative

If dp[target] is unchanged, no combination of coins can make that amount.

Which approach is commonly used to solve the Coin Change Minimum Coins problem efficiently?

A Brute Force Recursion without memoization

B Greedy algorithm always

C Sorting coins only

D Dynamic Programming

Dynamic programming efficiently builds solutions for all amounts up to the target.

What is the base case value for dp[0] in the Coin Change Minimum Coins problem?

A amount + 1

B 1

C 0

D -1

Zero coins are needed to make amount zero, so dp[0] = 0.

Explain how dynamic programming solves the Coin Change Minimum Coins problem step-by-step.

Think about building the solution from smaller amounts to larger amounts.

Describe what the dp array represents and how it changes during the Coin Change Minimum Coins algorithm.

Focus on the meaning of dp indices and values.

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)