Why Coin Change Minimum Coins in DSA Go?

The Big Idea

What if you could instantly know the fewest coins needed for any amount without guessing?

The Scenario

Imagine you have a pile of coins of different values and you need to pay a certain amount. You try to pick coins one by one, guessing which ones to use to make the exact amount with the fewest coins.

This guessing game can take forever if you do it by hand or with simple trial and error.

The Problem

Manually trying all combinations is slow and confusing. You might pick too many small coins or miss a better combination. It's easy to make mistakes and waste time.

Without a clear method, you can't be sure you found the smallest number of coins.

The Solution

The Coin Change Minimum Coins algorithm uses a smart step-by-step approach to find the fewest coins needed. It checks all possibilities quickly and remembers the best answers for smaller amounts to build up the solution.

This saves time and avoids mistakes, giving you the answer fast and correctly.

Before vs After

✗ Before

func minCoins(coins []int, amount int) int {
    // Try all combinations manually - very slow
    // No memory of past tries
    return -1 // placeholder
}

✓ After

func minCoins(coins []int, amount int) int {
    dp := make([]int, amount+1)
    for i := 1; i <= amount; i++ {
        dp[i] = amount + 1
        for _, coin := range coins {
            if coin <= i {
                if dp[i-coin]+1 < dp[i] {
                    dp[i] = dp[i-coin] + 1
                }
            }
        }
    }
    if dp[amount] > amount {
        return -1
    }
    return dp[amount]
}

What It Enables

This algorithm lets you quickly find the smallest number of coins needed to make any amount, saving time and effort.

Real Life Example

When a cashier needs to give change using the fewest coins possible, this method helps find the best way to do it fast and correctly.

Key Takeaways

Manual guessing is slow and error-prone.

The algorithm uses memory to build solutions efficiently.

It guarantees the minimum number of coins for any amount.

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