Coin Change Total Number of Ways in DSA Go

Mental Model

We want to find how many different ways we can make a total amount using given coin values. Each coin can be used many times.

Analogy: Imagine you have different types of Lego blocks (coins) and you want to build a tower of a certain height (amount). You count all the different ways to stack blocks to reach that height.

coins: [1, 2, 3]
amount: 4

Ways array (index = amount):
[1, 0, 0, 0, 0]
↑
Initially, 1 way to make 0 amount (use no coins).

Dry Run Walkthrough

Input: coins: [1, 2, 3], amount: 4

Goal: Find total number of ways to make amount 4 using coins 1, 2, and 3.

Step 1: Start with ways array: ways[0]=1 (one way to make 0), others 0

ways: [1, 0, 0, 0, 0]

Why: We know there is exactly one way to make amount 0: use no coins.

Step 2: Use coin 1: update ways for amounts 1 to 4 by adding ways[amount - 1]

ways: [1, 1, 1, 1, 1]

Why: For each amount, ways increase by ways to make amount minus coin 1.

Step 3: Use coin 2: update ways for amounts 2 to 4 by adding ways[amount - 2]

ways: [1, 1, 2, 2, 3]

Why: Add ways to make amount minus coin 2 to current ways.

Step 4: Use coin 3: update ways for amounts 3 to 4 by adding ways[amount - 3]

ways: [1, 1, 2, 3, 4]

Why: Add ways to make amount minus coin 3 to current ways.

Result:

ways: [1, 1, 2, 3, 4]
Total ways to make 4 = 4

Annotated Code

DSA Go

package main

import "fmt"

func coinChangeWays(coins []int, amount int) int {
    ways := make([]int, amount+1)
    ways[0] = 1 // one way to make amount 0: no coins

    for _, coin := range coins {
        for a := coin; a <= amount; a++ {
            ways[a] += ways[a-coin] // add ways to make (a - coin)
        }
    }

    return ways[amount]
}

func main() {
    coins := []int{1, 2, 3}
    amount := 4
    result := coinChangeWays(coins, amount)
    fmt.Printf("Total ways to make %d = %d\n", amount, result)
}

ways[0] = 1 // one way to make amount 0: no coins

initialize base case: one way to make zero amount

for _, coin := range coins {

iterate over each coin to update ways

for a := coin; a <= amount; a++ {

for each amount from coin value to target, update ways

ways[a] += ways[a-coin] // add ways to make (a - coin)

add ways to make amount minus current coin to current ways

Output

Total ways to make 4 = 4

Complexity Analysis

Time: O(n * amount) because we process each coin for all amounts up to target

Space: O(amount) because we store ways for all amounts from 0 to target

vs Alternative: Compared to recursive approach without memoization which is exponential, this dynamic programming approach is efficient and uses less time.

Edge Cases

amount = 0

Returns 1 because there is exactly one way to make zero amount: use no coins

DSA Go

ways[0] = 1 // one way to make amount 0: no coins

coins array empty

Returns 0 for any amount > 0 because no coins to use

DSA Go

for _, coin := range coins { ... } // loop skipped if coins empty

coins with duplicates

Processes each coin independently; duplicate values will overcount combinations treating them as distinct denominations (standard problem assumes unique denominations)

DSA Go

for _, coin := range coins { ... } // processes each coin independently

When to Use This Pattern

When asked to count the number of ways to make a sum using unlimited coins, use the coin change total number of ways pattern with dynamic programming.

Common Mistakes

Mistake: Updating ways array in wrong order causing overcounting
Fix: Iterate amounts from coin value up to target to avoid counting same combination multiple times

Summary

Counts how many ways to make a target amount using given coin denominations unlimited times.

Use when you need total combinations of coins to reach a sum, not just one solution.

The key is to build up ways from smaller amounts using each coin incrementally.

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