Coin Change Minimum Coins in DSA Go - Execution Trace

Concept Flow - Coin Change Minimum Coins

Start with amount

↓

Initialize dp array with max values

↓

Set dp[0

↓

For each coin

↓

For each amount from coin to target

↓

Update dp[amount

↓

After all coins processed

↓

Check dp[target

↓

If dp[target

↓

End

We build a dp array where each index represents the minimum coins needed for that amount, updating it by checking all coins.

Execution Sample

DSA Go

coins := []int{1, 2, 5}
amount := 11
dp := make([]int, amount+1)
for i := range dp { dp[i] = amount+1 }
dp[0] = 0
for _, coin := range coins {
  for x := coin; x <= amount; x++ {
    if dp[x-coin]+1 < dp[x] {
      dp[x] = dp[x-coin] + 1
    }
  }
}

This code finds the minimum number of coins needed to make amount 11 using coins 1, 2, and 5.

Execution Table

Step	Operation	Amount Index (x)	Coin Used	dp[x-coin]	dp[x]	Updated dp[x]	Visual dp Array State
1	Initialize dp	-	-	-	-	-	[12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12]
2	Set dp[0]	0	-	-	-	0	[0, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12]
3	Use coin 1	1	1	dp[0]=0	dp[1]=12	1	[0, 1, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12]
4	Use coin 1	2	1	dp[1]=1	dp[2]=12	2	[0, 1, 2, 12, 12, 12, 12, 12, 12, 12, 12, 12]
5	Use coin 1	3	1	dp[2]=2	dp[3]=12	3	[0, 1, 2, 3, 12, 12, 12, 12, 12, 12, 12, 12]
6	Use coin 1	4	1	dp[3]=3	dp[4]=12	4	[0, 1, 2, 3, 4, 12, 12, 12, 12, 12, 12, 12]
7	Use coin 1	5	1	dp[4]=4	dp[5]=12	5	[0, 1, 2, 3, 4, 5, 12, 12, 12, 12, 12, 12]
8	Use coin 1	6	1	dp[5]=5	dp[6]=12	6	[0, 1, 2, 3, 4, 5, 6, 12, 12, 12, 12, 12]
9	Use coin 1	7	1	dp[6]=6	dp[7]=12	7	[0, 1, 2, 3, 4, 5, 6, 7, 12, 12, 12, 12]
10	Use coin 1	8	1	dp[7]=7	dp[8]=12	8	[0, 1, 2, 3, 4, 5, 6, 7, 8, 12, 12, 12]
11	Use coin 1	9	1	dp[8]=8	dp[9]=12	9	[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 12]
12	Use coin 1	10	1	dp[9]=9	dp[10]=12	10	[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12]
13	Use coin 1	11	1	dp[10]=10	dp[11]=12	11	[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
14	Use coin 2	2	2	dp[0]=0	dp[2]=2	1	[0, 1, 1, 3, 4, 5, 6, 7, 8, 9, 10, 11]
15	Use coin 2	3	2	dp[1]=1	dp[3]=3	2	[0, 1, 1, 2, 4, 5, 6, 7, 8, 9, 10, 11]
16	Use coin 2	4	2	dp[2]=1	dp[4]=4	2	[0, 1, 1, 2, 2, 5, 6, 7, 8, 9, 10, 11]
17	Use coin 2	5	2	dp[3]=2	dp[5]=5	3	[0, 1, 1, 2, 2, 3, 6, 7, 8, 9, 10, 11]
18	Use coin 2	6	2	dp[4]=2	dp[6]=6	3	[0, 1, 1, 2, 2, 3, 3, 7, 8, 9, 10, 11]
19	Use coin 2	7	2	dp[5]=3	dp[7]=7	4	[0, 1, 1, 2, 2, 3, 3, 4, 8, 9, 10, 11]
20	Use coin 2	8	2	dp[6]=3	dp[8]=8	4	[0, 1, 1, 2, 2, 3, 3, 4, 4, 9, 10, 11]
21	Use coin 2	9	2	dp[7]=4	dp[9]=9	5	[0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 10, 11]
22	Use coin 2	10	2	dp[8]=4	dp[10]=10	5	[0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 11]
23	Use coin 2	11	2	dp[9]=5	dp[11]=11	6	[0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6]
24	Use coin 5	5	5	dp[0]=0	dp[5]=3	1	[0, 1, 1, 2, 2, 1, 3, 4, 4, 5, 5, 6]
25	Use coin 5	6	5	dp[1]=1	dp[6]=3	2	[0, 1, 1, 2, 2, 1, 2, 4, 4, 5, 5, 6]
26	Use coin 5	7	5	dp[2]=1	dp[7]=4	2	[0, 1, 1, 2, 2, 1, 2, 2, 4, 5, 5, 6]
27	Use coin 5	8	5	dp[3]=2	dp[8]=4	3	[0, 1, 1, 2, 2, 1, 2, 2, 3, 5, 5, 6]
28	Use coin 5	9	5	dp[4]=2	dp[9]=5	3	[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 5, 6]
29	Use coin 5	10	5	dp[5]=1	dp[10]=5	2	[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 6]
30	Use coin 5	11	5	dp[6]=2	dp[11]=6	3	[0, 1, 1, 2, 2, 1, 2, 2, 3, 3, 2, 3]

💡 dp[11] = 3 means minimum 3 coins needed for amount 11

Variable Tracker

Variable	Start	After Step 2	After Step 13	After Step 23	After Step 30 (Final)
dp	[12,12,12,12,12,12,12,12,12,12,12,12]	[0,12,12,12,12,12,12,12,12,12,12,12]	[0,1,2,3,4,5,6,7,8,9,10,11]	[0,1,1,2,2,3,3,4,4,5,5,6]	[0,1,1,2,2,1,2,2,3,3,2,3]
coin	-	1	1	2	5
x	-	-	11	11	11

Key Moments - 3 Insights

Why do we initialize dp array with amount+1 (a large number)?

We use amount+1 as a placeholder for 'infinity' because no amount can need more coins than amount itself. This helps us find minimum values correctly as shown in execution_table rows 1 and 2.

Why is dp[0] set to 0 at the start?

dp[0] = 0 means zero coins are needed to make amount 0. This is the base case that allows building solutions for larger amounts, as seen in execution_table row 2.

How does dp[x] get updated when using a coin?

For each amount x, if using coin c leads to fewer coins (dp[x-c] + 1 < dp[x]), we update dp[x]. This is shown in execution_table rows 3-30 where dp[x] values decrease step by step.

Visual Quiz - 3 Questions

Look at the execution_table at step 14, what is the updated dp[2] value after using coin 2?

A 1

B 2

C 12

D 0

💡 Hint

Check the 'Updated dp[x]' column at step 14 for amount index 2.

At which step does dp[5] first get updated to 1?

A Step 14

B Step 7

C Step 24

D Step 30

💡 Hint

Look for the first step where 'Updated dp[x]' for amount 5 changes to 1 in execution_table.

If we remove coin 1 from coins, how would dp[11] change at the end?

A It would remain 3

B It would increase or be unreachable

C It would become 0

D It would decrease

💡 Hint

Without coin 1, smaller amounts might not be reachable, check variable_tracker dp array final state.

Concept Snapshot

Coin Change Minimum Coins:
- Use dp array of size amount+1
- Initialize dp with amount+1 (infinity), dp[0]=0
- For each coin, update dp[x] = min(dp[x], dp[x-coin]+1)
- dp[amount] gives min coins or no solution if unchanged
- Bottom-up dynamic programming approach

Full Transcript

This visualization shows how to find the minimum number of coins needed to make a target amount using given coin denominations. We create a dp array where each index represents the minimum coins needed for that amount. Initially, dp is filled with a large number (amount+1) to represent unreachable states, except dp[0] which is zero because zero coins are needed for amount zero. We then iterate over each coin and for each amount from the coin value up to the target amount, we update dp[amount] if using the coin reduces the number of coins needed. The execution table traces each update step by step, showing how dp values improve. The final dp[target] value is the answer or indicates no solution if unchanged. Key moments clarify why dp is initialized this way, the importance of dp[0], and how updates happen. The quiz tests understanding of dp updates and effects of coin removal. This bottom-up dynamic programming method efficiently solves the coin change minimum coins problem.

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