Coin Change Total Number of Ways in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the ways array with the correct size.

DSA Go

ways := make([]int, [1])

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

Aamount - 1

Bamount

Clen(coins)

Damount + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using amount instead of amount + 1 causes index out of range errors.

Using length of coins instead of amount + 1.

The ways array should have a length of amount + 1 to include zero amount index.

2fill in blank

medium

Complete the code to set the base case for amount 0.

DSA Go

ways[0] = [1]

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

A0

B1

Camount

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting ways[0] to 0 means no ways to make amount 0, which is incorrect.

There is exactly one way to make amount 0: by choosing no coins.

3fill in blank

hard

Fix the error in the inner loop condition to correctly update ways.

DSA Go

for j := coin; j [1] amount + 1; j++ {
    ways[j] += ways[j - coin]
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index out of range error.

Using > or >= causes the loop to not run correctly.

The loop should run while j is less than amount + 1 to avoid index out of range.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA Go

func [1](coins []int, amount int) int {
    // function body
    return ways[[2]]
}

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

AcoinChangeWays

Bways

Camount

Dcoins

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing 'ways' or 'coins' for the index in ways[index].

Using incorrect function name.

The function is named coinChangeWays and returns ways[amount], the value in the ways array for the target amount.

5fill in blank

hard

Fill all three blanks to complete the nested loops for updating ways.

DSA Go

for _, [1] := range coins {
    for [2] := coin; [2] [3] amount + 1; [2]++ {
        ways[[2]] += ways[[2] - coin]
    }
}

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

Acoin

Bj

C<

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of coin in outer loop.

Using >= instead of < in inner loop condition.

The outer loop variable is coin, inner loop variable is j, and loop runs while j < amount + 1.

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