Coin Change Minimum Coins in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

dp := make([]int, amount+1)
for i := 1; i <= amount; i++ {
    dp[i] = [1]
}

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

A1

B-1

C0

Damount + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp[i] with 0 causes incorrect minimum calculations.

Using -1 can cause logic errors in comparisons.

We initialize dp[i] with amount+1 as a large number representing infinity, since the maximum coins needed can't exceed amount.

2fill in blank

medium

Complete the code to update dp[i] with the minimum coins needed.

DSA Go

for _, coin := range coins {
    if coin <= i {
        dp[i] = [1](dp[i], dp[i-coin]+1)
    }
}

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

Amax

Babs

Cmin

Dsum

Attempts:

3 left

💡 Hint

Common Mistakes

Using max instead of min leads to wrong results.

Forgetting to add 1 for the current coin.

We use min to keep the smallest number of coins needed for amount i.

3fill in blank

hard

Fix the error in the return statement to handle no solution case.

DSA Go

if dp[amount] == [1] {
    return -1
}
return dp[amount]

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

A0

Bamount + 1

C-1

Dmath.MaxInt32

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 0 instead of amount+1.

Returning dp[amount] without checking leads to wrong answers.

If dp[amount] is still amount+1, it means no combination was found, so return -1.

4fill in blank

hard

Fill both blanks to complete the loop that fills dp array.

DSA Go

for i := [1]; i <= amount; i++ {
    for _, coin := range [2] {
        if coin <= i {
            dp[i] = min(dp[i], dp[i-coin]+1)
        }
    }
}

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

A1

B0

Ccoins

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Starting i from 0 causes unnecessary checks.

Looping over amount instead of coins causes errors.

We start i from 1 to amount and iterate over coins to update dp.

5fill in blank

hard

Fill all three blanks to define the min function and call it correctly.

DSA Go

func min(a, b int) int {
    if [1] < [2] {
        return [3]
    }
    return b
}

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

Aa

Bb

Attempts:

3 left

💡 Hint

Common Mistakes

Returning b when a is smaller.

Swapping parameters in comparison.

The min function compares a and b and returns the smaller one.

Practice

1. What does the "Coin Change Minimum Coins" problem ask you to find?

easy

A. The smallest number of coins needed to make a given amount

B. The total number of different coins available

C. The largest coin value in the list

D. The sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of coins needed to make a specific amount using given coin denominations.

2. Step 2: Identify the correct description

   The smallest number of coins needed to make a given amount correctly describes this goal, while others describe unrelated concepts.

3. Final Answer:

   The smallest number of coins needed to make a given amount -> Option A

4. Quick Check:

   Minimum coins = smallest count needed [OK]

Hint: Focus on 'minimum coins' needed for the amount [OK]

Common Mistakes:

• Confusing total coin count with minimum coins
• Thinking it's about coin values sum
• Mixing largest coin with minimum coins

2. Which Go code snippet correctly initializes a slice to store minimum coins for amounts up to 5 with initial value 6 (assuming 6 means unreachable)?

easy

A. minCoins := make([]int, 6); for i := 0; i <= 6; i++ { minCoins[i] = 6 }

B. minCoins := []int{6,6,6,6,6}

C. minCoins := make([]int, 5); for i := 0; i < 5; i++ { minCoins[i] = 6 }

D. minCoins := make([]int, 6); for i := range minCoins { minCoins[i] = 6 }

Solution

1. Step 1: Check slice length and initialization

   We need a slice of length 6 (for amounts 0 to 5). minCoins := make([]int, 6); for i := range minCoins { minCoins[i] = 6 } creates a slice of length 6 and sets each element to 6 using a loop.

2. Step 2: Verify correctness of other options

   minCoins := []int{6,6,6,6,6} creates length 5 (too short); minCoins := make([]int, 5); for i := 0; i < 5; i++ { minCoins[i] = 6 } creates length 5 (too short); minCoins := make([]int, 6); for i := 0; i <= 6; i++ { minCoins[i] = 6 } loops out of range (index 6 invalid).

3. Final Answer:

   minCoins := make([]int, 6); for i := range minCoins { minCoins[i] = 6 } -> Option D

4. Quick Check:

   Slice length 6 and loop within range [OK]

Hint: Use make with length and loop with range for safe init [OK]

Common Mistakes:

• Creating slice with wrong length
• Looping beyond slice length causing runtime error
• Using literal slice with insufficient length

3. What is the output of the following Go code snippet?

coins := []int{1, 3, 4}
amount := 6
minCoins := make([]int, amount+1)
for i := 1; i <= amount; i++ {
    minCoins[i] = amount + 1
}
minCoins[0] = 0
for i := 1; i <= amount; i++ {
    for _, coin := range coins {
        if coin <= i && minCoins[i-coin]+1 < minCoins[i] {
            minCoins[i] = minCoins[i-coin] + 1
        }
    }
}
fmt.Println(minCoins[amount])

medium

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Initialize minCoins array

   minCoins starts as [0,7,7,7,7,7,7] (7 = amount+1). We update minCoins for each amount from 1 to 6.

2. Step 2: Calculate minimum coins for amount 6

   Check coins 1,3,4 for each i:
   - i=1: minCoins[1]=minCoins[0]+1=1 (coin 1)
   - i=2: minCoins[2]=minCoins[1]+1=2 (coin 1)
   - i=3: minCoins[3]=minCoins[0]+1=1 (coin 3)
   - i=4: minCoins[4]=minCoins[0]+1=1 (coin 4)
   - i=5: minCoins[5]=minCoins[4]+1=2 (coin 1)
   - i=6: minCoins[6]=minCoins[3]+1=2 (coin 3)

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Minimum coins for 6 = 2 (e.g., 3+3) [OK]

Hint: Trace inner loop updates for each amount [OK]

Common Mistakes:

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

4. Find the error in this Go code for minimum coin change and select the fix:

coins := []int{1, 2, 5}
amount := 11
minCoins := make([]int, amount+1)
for i := 1; i <= amount; i++ {
    minCoins[i] = amount
}
minCoins[0] = 0
for i := 1; i <= amount; i++ {
    for _, coin := range coins {
        if coin < i && minCoins[i-coin]+1 < minCoins[i] {
            minCoins[i] = minCoins[i-coin] + 1
        }
    }
}
fmt.Println(minCoins[amount])

medium

A. Change 'coin < i' to 'coin <= i' in the if condition

B. Initialize minCoins with zeros instead of amount

C. Set minCoins[0] = amount instead of 0

D. Change outer loop to 'for i := 0; i < amount; i++'

Solution

1. Step 1: Identify the condition error

   The condition 'coin < i' skips cases where coin equals i, missing exact matches (e.g., coin=5, i=5).

2. Step 2: Correct the condition

   Changing to 'coin <= i' includes exact matches, allowing minCoins[i] to update correctly.

3. Final Answer:

   Change 'coin < i' to 'coin <= i' in the if condition -> Option A

4. Quick Check:

   Include coin equal to amount index [OK]

Hint: Use <= to include exact coin matches [OK]

Common Mistakes:

• Using < instead of <= causing missed updates
• Wrong initialization of minCoins array
• Incorrect loop ranges causing index errors

5. Given coins = [1, 5, 10, 25] and amount = 30, what is the minimum number of coins needed to make 30 using the coin change minimum coins algorithm?

medium

A. 4

B. 3

C. 2

D. 5

Solution

1. Step 1: Analyze coin denominations and amount

   Coins are 1, 5, 10, 25. Amount is 30.

2. Step 2: Find minimum coins to make 30

   Use one 25 coin + one 5 coin = 2 coins total. Using smaller coins would require more coins.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   25 + 5 = 30 with 2 coins [OK]

Hint: Use largest coins first to minimize count [OK]

Common Mistakes:

• Assuming only 1-coin denominations
• Not considering combinations with larger coins
• Counting coins incorrectly