Coin Change Total Number of Ways in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of the following Go code that calculates the total number of ways to make change for amount 5 using coins [1, 2, 3]?

DSA Go

package main
import "fmt"
func coinChangeWays(coins []int, amount int) int {
    dp := make([]int, amount+1)
    dp[0] = 1
    for _, coin := range coins {
        for i := coin; i <= amount; i++ {
            dp[i] += dp[i-coin]
        }
    }
    return dp[amount]
}
func main() {
    coins := []int{1, 2, 3}
    amount := 5
    fmt.Println(coinChangeWays(coins, amount))
}

A 5

B 6

C 7

D 8

💡 Hint

Think about how many combinations of coins can sum up to 5 using 1, 2, and 3.

The code uses dynamic programming to count the number of ways to make change. For amount 5 and coins [1, 2, 3], the total ways are 7.

❓ Predict Output

intermediate

Number of Ways to Make Change for Amount 4

What does this Go function print when called with coins [2, 5, 3, 6] and amount 4?

DSA Go

package main
import "fmt"
func coinChangeWays(coins []int, amount int) int {
    dp := make([]int, amount+1)
    dp[0] = 1
    for _, coin := range coins {
        for i := coin; i <= amount; i++ {
            dp[i] += dp[i-coin]
        }
    }
    return dp[amount]
}
func main() {
    coins := []int{2, 5, 3, 6}
    amount := 4
    fmt.Println(coinChangeWays(coins, amount))
}

A 2

B 0

C 1

D 3

💡 Hint

Only coins 2 and 3 can be used to make 4, check combinations.

The ways to make 4 are using two 2-coin pieces or one 3-coin plus one 1-coin (but 1 is not in coins), so only two 2-coin pieces and one 3-coin plus one 1-coin is invalid. Actually, only two 2-coin pieces and one 3-coin plus one 1-coin is invalid. So total ways is 1.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error will this Go code produce when run, and why?

DSA Go

package main
import "fmt"
func coinChangeWays(coins []int, amount int) int {
    dp := make([]int, amount)
    dp[0] = 1
    for _, coin := range coins {
        for i := coin; i <= amount; i++ {
            dp[i] += dp[i-coin]
        }
    }
    return dp[amount]
}
func main() {
    coins := []int{1, 2}
    amount := 3
    fmt.Println(coinChangeWays(coins, amount))
}

A panic: runtime error: index out of range

B 0

C 3

D compile error: cannot assign to dp[0]

💡 Hint

Check the size of dp slice and the indices accessed.

The dp slice is created with length amount (3), but indices go up to amount (3), which is out of range since indexing is 0-based.

❓ Predict Output

advanced

Output of Coin Change with Zero Amount

What is the output of this Go program when amount is zero?

DSA Go

package main
import "fmt"
func coinChangeWays(coins []int, amount int) int {
    dp := make([]int, amount+1)
    dp[0] = 1
    for _, coin := range coins {
        for i := coin; i <= amount; i++ {
            dp[i] += dp[i-coin]
        }
    }
    return dp[amount]
}
func main() {
    coins := []int{1, 2, 3}
    amount := 0
    fmt.Println(coinChangeWays(coins, amount))
}

A 0

B panic: runtime error

C nil

D 1

💡 Hint

Think about how many ways to make zero amount.

There is exactly one way to make amount zero: use no coins.

🧠 Conceptual

expert

Why Does the Coin Change DP Use Outer Loop Over Coins?

In the coin change total number of ways problem, why is the outer loop iterating over coins and the inner loop over amounts, rather than the other way around?

A To optimize memory usage by limiting dp array size

B To ensure combinations are counted without duplicates by processing coins in order

C To allow using each coin unlimited times in any order

D To reduce the time complexity from exponential to linear

💡 Hint

Think about how order of loops affects counting unique combinations.

Iterating coins first ensures combinations are counted without duplicates because each coin is considered in sequence, preventing counting the same combination multiple times in different orders.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total number of ways to make a target amount using given coin denominations.

B. The minimum number of coins needed to make a target amount.

C. The maximum value of coins that can be used to reach the target.

D. The total number of coins available in the input.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible combinations of coins that sum up to the target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts all ways, not the least number of coins.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

4. Quick Check:

   Counting combinations = total ways [OK]

Hint: Count combinations, not minimum coins, for total ways [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking about maximum coin value instead of count
• Assuming order of coins matters

2. Which Go code snippet correctly initializes the DP array for the Coin Change Total Number of Ways problem for amount 5?

easy

A. dp := make([]int, 6); dp[0] = 1

B. dp := make([]int, 5); dp[0] = 0

C. dp := make([]int, 5); dp[0] = 1

D. dp := make([]int, 6); dp[0] = 0

Solution

1. Step 1: Determine DP array size

   For amount 5, dp array size should be 6 to include index 0 to 5.

2. Step 2: Initialize base case

   dp[0] = 1 because there is exactly one way to make amount 0 (using no coins).

3. Final Answer:

   dp := make([]int, 6); dp[0] = 1 -> Option A

4. Quick Check:

   dp size = amount+1 and dp[0] = 1 [OK]

Hint: DP array size = amount+1; dp[0] = 1 base case [OK]

Common Mistakes:

• Using dp array size equal to amount instead of amount+1
• Setting dp[0] to 0 instead of 1
• Forgetting to initialize dp[0]

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

coins := []int{1, 2}
amount := 3
dp := make([]int, amount+1)
dp[0] = 1
for _, coin := range coins {
    for x := coin; x <= amount; x++ {
        dp[x] += dp[x-coin]
    }
}
fmt.Println(dp[amount])

medium

A. 4

B. 2

C. 3

D. 1

Solution

1. Step 1: Initialize dp array and base case

   dp = [1,0,0,0] initially; dp[0]=1 means one way to make 0.

2. Step 2: Process coin 1

   For x=1 to 3, dp[x] += dp[x-1]: dp[1]=1, dp[2]=1, dp[3]=1.

3. Step 3: Process coin 2

   For x=2 to 3, dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2.

4. Step 4: Final dp array

   dp = [1,1,2,2], so dp[3] = 2.

5. Final Answer:

   2 -> Option B

6. Quick Check:

   Ways to make 3 with coins 1,2 = 2 [OK]

Hint: Add ways from smaller amounts for each coin [OK]

Common Mistakes:

• Confusing dp[x] update order
• Not including dp[0] base case
• Miscounting combinations for amount 3

4. Identify the bug in this Go code for Coin Change Total Number of Ways:

coins := []int{1, 2, 3}
amount := 4
dp := make([]int, amount)
dp[0] = 1
for _, coin := range coins {
    for x := coin; x <= amount; x++ {
        dp[x] += dp[x-coin]
    }
}
fmt.Println(dp[amount])

medium

A. The coins slice should include 0 as a coin

B. dp[0] should be 0, not 1

C. dp array size should be amount+1, not amount

D. The inner loop should start from 0, not coin

Solution

1. Step 1: Check dp array size

   dp is created with size amount (4), but indices go up to amount (4), so dp[4] access is out of range.

2. Step 2: Correct dp size and base case

   dp should be size amount+1 (5) to include dp[4], and dp[0] = 1 is correct.

3. Final Answer:

   dp array size should be amount+1, not amount -> Option C

4. Quick Check:

   dp size = amount+1 to avoid index error [OK]

Hint: dp size must be amount+1 to cover all indices [OK]

Common Mistakes:

• Using dp size equal to amount causing index out of range
• Starting inner loop incorrectly
• Setting dp[0] to 0 instead of 1

5. Given coins = []int{1, 3, 5} and amount = 7, how many total ways are there to make amount 7 using these coins? Use the Coin Change Total Number of Ways approach.

hard

A. 7

B. 5

C. 6

D. 4

Solution

1. Step 1: Initialize dp array

   dp size = 8 (0 to 7), dp[0] = 1, others 0.

2. Step 2: Process coin 1

   Update dp for amounts 1 to 7 by adding dp[x-1]: dp = [1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   Update dp for x=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3

4. Step 4: Process coin 5

   Update dp for x=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4

5. Step 5: Final dp array

   dp = [1,1,1,2,2,3,4,4], so dp[7] = 4.

6. Final Answer:

   4 -> Option D

7. Quick Check:

   Count ways with coins 1,3,5 for amount 7 = 4 [OK]

Hint: Update dp for each coin sequentially to count ways [OK]

Common Mistakes:

• Not updating dp correctly for each coin
• Mixing order of coin processing
• Miscounting combinations for amount 7