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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Minimum Coins Calculation

What is the output of this Go code that calculates the minimum number of coins needed to make change for 11 using coins [1, 2, 5]?

DSA Go

package main
import "fmt"
func minCoins(coins []int, amount int) int {
    dp := make([]int, amount+1)
    for i := 1; i <= amount; i++ {
        dp[i] = amount + 1
        for _, c := range coins {
            if c <= i && dp[i-c]+1 < dp[i] {
                dp[i] = dp[i-c] + 1
            }
        }
    }
    if dp[amount] > amount {
        return -1
    }
    return dp[amount]
}
func main() {
    coins := []int{1, 2, 5}
    amount := 11
    fmt.Println(minCoins(coins, amount))
}

A 5

B 2

C 3

D -1

💡 Hint

Think about the fewest coins that add up to 11 using 1, 2, and 5.

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

❓ Predict Output

intermediate

Minimum Coins for Amount Not Possible

What is the output of this Go code when trying to make change for 3 using coins [2, 4]?

DSA Go

package main
import "fmt"
func minCoins(coins []int, amount int) int {
    dp := make([]int, amount+1)
    for i := 1; i <= amount; i++ {
        dp[i] = amount + 1
        for _, c := range coins {
            if c <= i && dp[i-c]+1 < dp[i] {
                dp[i] = dp[i-c] + 1
            }
        }
    }
    if dp[amount] > amount {
        return -1
    }
    return dp[amount]
}
func main() {
    coins := []int{2, 4}
    amount := 3
    fmt.Println(minCoins(coins, amount))
}

A -1

B 1

C 0

D 3

💡 Hint

Check if the amount can be formed by any combination of coins.

3 cannot be formed by any combination of 2 and 4, so the function returns -1.

🔧 Debug

advanced

Identify the Bug in Coin Change Code

This Go code is supposed to find the minimum coins needed for amount 7 with coins [1, 3, 4]. What is the bug causing wrong output?

DSA Go

package main
import "fmt"
func minCoins(coins []int, amount int) int {
    dp := make([]int, amount+1)
    dp[0] = 0
    for i := 1; i <= amount; i++ {
        dp[i] = amount
        for _, c := range coins {
            if c <= i && dp[i-c]+1 < dp[i] {
                dp[i] = dp[i-c] + 1
            }
        }
    }
    if dp[amount] > amount {
        return -1
    }
    return dp[amount]
}
func main() {
    coins := []int{1, 3, 4}
    amount := 7
    fmt.Println(minCoins(coins, amount))
}

A dp array initialization should use amount+1 instead of amount

B dp[0] is not initialized to 0, causing wrong results

C The loop should start from 0 instead of 1

D The condition c <= i should be c < i

💡 Hint

Check the base case initialization for dp array.

dp[0] must be 0 because zero coins are needed to make amount 0. Without this, dp[i-c] uses zero value 0 incorrectly.

🧠 Conceptual

advanced

Time Complexity of Coin Change Minimum Coins

What is the time complexity of the dynamic programming solution for minimum coins problem with n coins and amount m?

A O(n^m)

B O(n + m)

C O(m^n)

D O(n * m)

💡 Hint

Consider the nested loops over amount and coins.

The solution uses a loop over amount (m) and inside it loops over coins (n), so total O(n*m).

🚀 Application

expert

Minimum Coins for Large Amount with Limited Coins

Given coins [1, 5, 10, 25] and amount 63, what is the minimum number of coins needed?

A 6

B 7

C 5

D 4

💡 Hint

Try to use the largest coins first to minimize count.

63 = 25 + 25 + 10 + 1 + 1 + 1 = 6 coins total.

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