Coin Change Total Number of Ways in DSA Javascript - 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 make a certain amount of money using these coins.Imagine you have coins like 1, 2, and 5 cents, and you want to make 5 cents. How many different combinations of these coins can you use?

🎯 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 an array called coins with the values 1, 2, and 5

Create a variable called targetAmount and set it to 5

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

Use a dynamic programming approach with a 1D array to count the ways

Print the total number of ways returned by countWays(coins, targetAmount)

💡 Why This Matters

🌍 Real World

This problem is similar to figuring out how many ways you can pay a bill using different coins or bills in your wallet.

💼 Career

Understanding dynamic programming and counting combinations is useful in software development, especially in optimization problems and algorithm design.

1

Create the coins array

Create an array called coins with the values 1, 2, and 5

DSA Javascript

# Your code here

Hint

Use square brackets [] to create an array and separate values with commas.

2

Set the target amount

Create a variable called targetAmount and set it to 5

DSA Javascript

const coins = [1, 2, 5];
# Your code here

Hint

Use const to create a variable that does not change.

3

Write the countWays function

Write a function called countWays that takes coins and targetAmount as inputs and returns the total number of ways to make the amount using a 1D array for dynamic programming. Use a for loop with variable coin to go through coins and a nested for loop with variable amount from coin to targetAmount. Update the ways array inside the loops.

DSA Javascript

const coins = [1, 2, 5];
const targetAmount = 5;

# Your code here

Hint

Initialize an array ways with zeros and set ways[0] to 1. Then use nested loops to update ways.

4

Print the total number of ways

Print the total number of ways by calling countWays(coins, targetAmount) using console.log

DSA Javascript

const coins = [1, 2, 5];
const targetAmount = 5;

function countWays(coins, targetAmount) {
    const ways = Array(targetAmount + 1).fill(0);
    ways[0] = 1;
    for (const coin of coins) {
        for (let amount = coin; amount <= targetAmount; amount++) {
            ways[amount] += ways[amount - coin];
        }
    }
    return ways[targetAmount];
}

# Your code here

Hint

Use console.log(countWays(coins, targetAmount)) to print the result.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?
coins = [1, 2, 3], amount = 4

easy

A. The maximum coin value used to make the amount

B. The minimum number of coins needed to make the amount

C. The total number of ways to make the amount using given coins

D. The sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for how many different ways we can combine coins to reach the target amount.

2. Step 2: Differentiate from similar problems

   This is not about minimum coins or maximum coin value, but counting all possible combinations.

3. Final Answer:

   The total number of ways to make the amount using given coins -> Option C

4. Quick Check:

   Counting combinations = total ways [OK]

Hint: Count combinations, not minimum or max coin values [OK]

Common Mistakes:

• Confusing with minimum coin count problem
• Thinking it asks for max coin value
• Summing coin values instead of counting ways

2. Which of the following is the correct way to initialize the DP array for the Coin Change Total Number of Ways problem in JavaScript?

const amount = 5;
const dp = ...;

easy

A. new Array(amount + 1).fill(0); dp[0] = 1;

B. new Array(amount).fill(1); dp[0] = 0;

C. new Array(amount + 1).fill(1); dp[0] = 0;

D. new Array(amount).fill(0); dp[0] = 1;

Solution

1. Step 1: Initialize DP array size

   We need dp array of size amount + 1 to include zero amount index.

2. Step 2: Set base case

   dp[0] = 1 because there is exactly one way to make amount 0 (use no coins).

3. Final Answer:

   new Array(amount + 1).fill(0); dp[0] = 1; -> Option A

4. Quick Check:

   dp size = amount+1, dp[0] = 1 [OK]

Hint: DP array size = amount+1, dp[0] = 1 base case [OK]

Common Mistakes:

• Using array size = amount instead of amount+1
• Setting dp[0] = 0 instead of 1
• Filling dp with 1s initially

3. What is the output of the following code?

const coins = [1, 2];
const amount = 3;
const dp = new Array(amount + 1).fill(0);
dp[0] = 1;
for (const coin of coins) {
  for (let x = coin; x <= amount; x++) {
    dp[x] += dp[x - coin];
  }
}
console.log(dp[amount]);

medium

A. 1

B. 2

C. 3

D. 0

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

   Update dp for amounts 1 to 3:
   dp[1]+=dp[0]=1, dp[2]+=dp[1]=1, dp[3]+=dp[2]=1
   dp = [1,1,1,1]

3. Step 3: Process coin 2

   Update dp for amounts 2 to 3:
   dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2
   dp = [1,1,2,2]

4. Final Answer:

   2 -> Option B

5. Quick Check:

   Ways to make 3 with [1,2] = 2 [OK]

Hint: Add ways from smaller amounts using each coin [OK]

Common Mistakes:

• Confusing dp update order
• Not initializing dp[0] = 1
• Miscounting combinations

4. Find the error in this code snippet for Coin Change Total Number of Ways:

const coins = [1, 2, 3];
const amount = 4;
const dp = new Array(amount + 1).fill(0);
dp[0] = 0;
for (const coin of coins) {
  for (let x = coin; x <= amount; x++) {
    dp[x] += dp[x - coin];
  }
}
console.log(dp[amount]);

medium

A. dp array size is incorrect

B. Coins array should be sorted

C. Loop should start from 0, not coin

D. dp[0] should be 1, not 0

Solution

1. Step 1: Check dp initialization

   dp[0] = 0 is wrong because base case means one way to make amount 0 (no coins).

2. Step 2: Correct base case

   Set dp[0] = 1 to represent one way to make zero amount.

3. Final Answer:

   dp[0] should be 1, not 0 -> Option D

4. Quick Check:

   Base case dp[0] = 1 [OK]

Hint: Always set dp[0] = 1 for base case [OK]

Common Mistakes:

• Setting dp[0] = 0
• Starting inner loop incorrectly
• Ignoring base case importance

5. Given coins = [1, 3, 5] and amount = 7, how many total ways are there to make amount 7?
Use the Coin Change Total Number of Ways approach.

hard

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0,0,0,0,0] with dp[0] = 1.

2. Step 2: Process coin 1

   Update dp for amounts 1 to 7:
   dp[1]=1, dp[2]=1, dp[3]=1, dp[4]=1, dp[5]=1, dp[6]=1, dp[7]=1

3. Step 3: Process coin 3

   Update dp for amounts 3 to 7:
   dp[3]+=dp[0]=1+1=2,
   dp[4]+=dp[1]=1+1=2,
   dp[5]+=dp[2]=1+1=2,
   dp[6]+=dp[3]=1+2=3,
   dp[7]+=dp[4]=1+2=3

4. Step 4: Process coin 5

   Update dp for amounts 5 to 7:
   dp[5]+=dp[0]=2+1=3,
   dp[6]+=dp[1]=3+1=4,
   dp[7]+=dp[2]=3+1=4

5. Final Answer:

   4 -> Option A

6. Quick Check:

   Ways to make 7 with [1,3,5] = 4 [OK]

Hint: Update dp for each coin from coin to amount [OK]

Common Mistakes:

• Not updating dp correctly for each coin
• Forgetting to add previous dp values
• Miscounting combinations