Coin Change Minimum Coins in DSA Go - Time & Space Complexity

Time Complexity: Coin Change Minimum Coins

O(amount x number_of_coins)

Understanding Time Complexity

We want to know how the time needed to find the minimum coins changes as the amount grows.

How does the algorithm's work increase when the amount or coin types increase?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 {
                dp[i] = min(dp[i], dp[i-coin]+1)
            }
        }
    }
    if dp[amount] > amount {
        return -1
    }
    return dp[amount]
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

This code finds the fewest coins needed to make a given amount using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops - outer loop runs for each amount from 1 to amount, inner loop runs over all coin types.
• How many times: Outer loop runs amount times, inner loop runs len(coins) times for each outer iteration.

How Execution Grows With Input

As the amount grows, the number of operations grows roughly by amount times number of coins.

Input Size (amount)	Approx. Operations
10	10 x number_of_coins
100	100 x number_of_coins
1000	1000 x number_of_coins

Pattern observation: Operations increase linearly with the amount and linearly with the number of coin types.

Final Time Complexity

Time Complexity: O(amount x number_of_coins)

This means the time needed grows proportionally with both the amount and how many coin types we have.

Common Mistake

[X] Wrong: "The algorithm runs in constant time because it just loops once."

[OK] Correct: The algorithm has two loops: one over the amount and one over the coins, so time grows with both.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves over naive methods and shows your grasp of nested loops and their impact.

Self-Check

"What if we sorted the coins and stopped the inner loop early when coin > current amount? How would the time complexity change?"

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