Coin Change Total Number of Ways in DSA Go - Build from Scratch

Coin Change Total Number of Ways

📖 Scenario: You have a collection of coins with different values. You want to find out how many different ways you can combine these coins to reach a certain amount of money.Imagine you are at a store and want to pay exactly 5 units using coins of 1, 2, and 3 units. How many ways can you do this?

🎯 Goal: Build a program that calculates the total number of ways to make a target amount using given coin values.

📋 What You'll Learn

Create a slice called coins with the exact values 1, 2, and 3

Create an integer variable called target and set it to 5

Write a function called countWays that takes coins and target as parameters and returns the total number of ways to make the target amount

Print the result of countWays(coins, target)

💡 Why This Matters

🌍 Real World

This problem models how to count ways to make change, useful in finance, vending machines, and resource allocation.

💼 Career

Understanding dynamic programming and combinatorial counting is important for software engineers working on optimization, algorithms, and problem-solving.

1

Create the coin values slice

Create a slice called coins with the exact integer values 1, 2, and 3.

DSA Go

package main

func main() {
    // Create the coins slice with values 1, 2, and 3
    // Your code here
}

Hint

Use coins := []int{1, 2, 3} to create the slice.

2

Set the target amount

Create an integer variable called target and set it to 5.

DSA Go

package main

func main() {
    coins := []int{1, 2, 3}
    // Create the target variable and set it to 5
    // Your code here
}

Hint

Use target := 5 to set the target amount.

3

Write the countWays function

Write a function called countWays that takes a slice of integers called coins and an integer target as parameters and returns an integer representing the total number of ways to make the target amount using the coins. Use a dynamic programming approach with a slice called dp where dp[i] stores the number of ways to make amount i. Initialize dp[0] to 1. For each coin, update the dp slice accordingly.

DSA Go

package main

func main() {
    coins := []int{1, 2, 3}
    target := 5

    // Write the countWays function below
    // Your code here
}

Hint

Use two nested loops: outer over coins, inner over amounts from coin to target. Update dp[amount] by adding dp[amount - coin].

4

Print the total number of ways

Print the result of calling countWays(coins, target) inside the main function.

DSA Go

package main

import "fmt"

func countWays(coins []int, target int) int {
    dp := make([]int, target+1)
    dp[0] = 1

    for _, coin := range coins {
        for amount := coin; amount <= target; amount++ {
            dp[amount] += dp[amount-coin]
        }
    }

    return dp[target]
}

func main() {
    coins := []int{1, 2, 3}
    target := 5

    // Print the total number of ways to make the target
    // Your code here
}

Hint

Use fmt.Println(countWays(coins, target)) to print the result.

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