Coin Change Total Number of Ways in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'Coin Change Total Number of Ways' problem ask you to find?

It asks how many different ways you can make a certain amount using given coin denominations, where order does not matter.

beginner

In the coin change problem, why do we use dynamic programming?

Because it helps us build solutions for smaller amounts and reuse them to find the total ways for bigger amounts efficiently.

intermediate

What does the dp array represent in the coin change total ways solution?

dp[i] represents the number of ways to make the amount i using the given coins.

intermediate

Why do we update dp[j] by adding dp[j - coin] in the solution?

Because dp[j - coin] is the number of ways to make amount j - coin, so adding coin extends those ways to amount j.

beginner

What is the initial value of dp[0] and why?

dp[0] is 1 because there is exactly one way to make amount 0 — by choosing no coins.

What does dp[i] represent in the coin change total number of ways problem?

A Minimum coins needed to make amount i

B Number of ways to make amount i

C Maximum coins used to make amount i

D Number of coins available

dp[i] stores the number of ways to make amount i using the given coins.

Why is dp[0] initialized to 1 in the coin change problem?

A Because amount 0 can be made in one way: no coins

B Because amount 0 requires one coin

C Because dp array must start with 1

D Because 0 is the smallest coin

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

When updating dp[j], why do we add dp[j - coin]?

A To reset dp[j]

B To remove duplicate ways

C To find minimum coins

D To count new ways by adding the current coin

Adding dp[j - coin] counts all ways to make amount j by including the current coin.

If coins = [1, 2] and amount = 3, how many ways are there to make 3?

A 3

B 1

C 2

D 4

Ways: (1+1+1), (1+2), (2+1) but order does not matter so (1+2) and (2+1) count as one way, total 2 ways.

Which approach is commonly used to solve the coin change total number of ways problem?

A Dynamic Programming

B Greedy Algorithm

C Divide and Conquer

D Backtracking

Dynamic Programming efficiently counts ways by building up from smaller amounts.

Explain how dynamic programming helps solve the coin change total number of ways problem.

Think about how you count ways for smaller amounts first.

Describe the role of the dp array and how it is updated in the coin change total number of ways solution.

Focus on what dp[i] means and how adding coins changes it.

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