Coin Change Total Number of Ways in DSA Go - Execution Trace

Concept Flow - Coin Change Total Number of Ways

Start with ways array initialized to 0 except ways[0

↓

For each coin in coins

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Result is ways[target

We build up the number of ways to make each amount by adding ways contributed by each coin step by step.

Execution Sample

DSA Go

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

Calculate total ways to make amount 4 using coins 1, 2, and 3.

Execution Table

Step	Operation	Amount Index (x)	ways[x-coin]	ways[x] Before	ways[x] After	Visual ways Array
1	Initialize ways[0]	-	-	0	1	[1,0,0,0,0]
2	Use coin=1, x=1	1	ways[0]=1	0	1	[1,1,0,0,0]
3	Use coin=1, x=2	2	ways[1]=1	0	1	[1,1,1,0,0]
4	Use coin=1, x=3	3	ways[2]=1	0	1	[1,1,1,1,0]
5	Use coin=1, x=4	4	ways[3]=1	0	1	[1,1,1,1,1]
6	Use coin=2, x=2	2	ways[0]=1	1	2	[1,1,2,1,1]
7	Use coin=2, x=3	3	ways[1]=1	1	2	[1,1,2,2,1]
8	Use coin=2, x=4	4	ways[2]=2	1	3	[1,1,2,2,3]
9	Use coin=3, x=3	3	ways[0]=1	2	3	[1,1,2,3,3]
10	Use coin=3, x=4	4	ways[1]=1	3	4	[1,1,2,3,4]
11	End	-	-	-	-	[1,1,2,3,4]

💡 All coins processed, final ways[4] = 4 means 4 ways to make amount 4.

Variable Tracker

Variable	Start	After Step 5	After Step 8	After Step 10	Final
ways	[1,0,0,0,0]	[1,1,1,1,1]	[1,1,2,2,3]	[1,1,2,3,4]	[1,1,2,3,4]
coin	-	1	2	3	3
x	-	-	4	4	-

Key Moments - 3 Insights

Why do we start ways[0] as 1 instead of 0?

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This is the base case needed to build up other amounts, as seen in step 1 of the execution_table.

Why do we add ways[x - coin] to ways[x] instead of replacing it?

We add because each coin can contribute additional ways to make amount x without removing previous ways. This cumulative addition is shown in steps 6-10 where ways[x] increases step by step.

Why do we loop coins first, then amounts, not the other way?

Looping coins first ensures combinations are counted without duplicates. If we looped amounts first, we might count the same combination multiple times. The execution_table rows 2-10 show how each coin updates ways progressively.

Visual Quiz - 3 Questions

Look at the execution_table at step 8, what is ways[4] after updating with coin 2?

A 3

B 1

C 2

D 4

💡 Hint

Check the 'ways[x] After' column for step 8 in execution_table.

At which step does ways[3] first become 3?

A Step 5

B Step 9

C Step 8

D Step 10

💡 Hint

Look at the 'ways[x] After' column for ways[3] in execution_table rows.

If we remove coin 1 from coins, what would ways[1] be after processing all coins?

A 2

B 1

C 0

D 4

💡 Hint

Without coin 1, ways to make amount 1 is zero since no smaller coin exists; check variable_tracker for ways array start.

Concept Snapshot

Coin Change Total Number of Ways:
- Initialize ways array with ways[0]=1
- For each coin, update ways[x] += ways[x - coin] for x >= coin
- ways[amount] holds total ways to make amount
- Order of loops avoids counting duplicates
- Dynamic programming builds solution bottom-up

Full Transcript

This visualization shows how to find the total number of ways to make a target amount using given coin denominations. We start with an array 'ways' where ways[0] is set to 1, meaning one way to make zero amount. For each coin, we update the ways array by adding the number of ways to make the current amount minus the coin value. This process accumulates all possible combinations without duplicates. The final answer is ways[amount]. The execution table traces each update step by step, showing how the array changes. Key points include why ways[0] is 1, why we add instead of replace, and why the coin loop is outer. The quiz tests understanding of these updates and the effect of removing coins.

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