Coin Change Minimum Coins in DSA Python - Time & Space Complexity

Time Complexity: Coin Change Minimum Coins

O(amount x number_of_coins)

Understanding Time Complexity

We want to know how the time needed to find the minimum coins changes as the amount grows.

How does the number of steps grow when we try to make bigger amounts with given coins?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the fewest coins needed to make the given amount using the coin list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops - outer loop runs through amounts from 1 to amount, inner loop runs through each coin.
• How many times: Outer loop runs amount times, inner loop runs len(coins) times for each outer iteration.

How Execution Grows With Input

As the amount grows, the code checks each coin for every smaller amount up to the target.

Input Size (amount)	Approx. Operations
10	10 x number_of_coins
100	100 x number_of_coins
1000	1000 x number_of_coins

Pattern observation: Operations grow roughly in direct proportion to the amount and number of coins.

Final Time Complexity

Time Complexity: O(amount x number_of_coins)

This means the time needed grows linearly with the amount and the number of coin types.

Common Mistake

[X] Wrong: "The time grows only with the amount, coins don't affect it much."

[OK] Correct: Each coin is checked for every amount, so more coins mean more checks and more time.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solves problems efficiently by breaking them into smaller parts.

Self-Check

"What if we used recursion with memoization instead of a bottom-up loop? How would the time complexity change?"

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)