Coin Change Total Number of Ways in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the ways array with 1 at index 0.

DSA Javascript

const ways = new Array(amount + 1).fill(0);
ways[[1]] = 1;

Drag options to blanks, or click blank then click option'

A0

B1

Camount

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting ways[1] = 1 instead of ways[0] = 1

Initializing all elements to 1

Forgetting to initialize ways array

The number of ways to make amount 0 is always 1 (using no coins), so ways[0] = 1.

2fill in blank

medium

Complete the code to loop through each coin in the coins array.

DSA Javascript

for (let [1] = 0; [1] < coins.length; [1]++) {
  const coin = coins[[1]];
  // rest of the code
}

Drag options to blanks, or click blank then click option'

Aindex

Bcoin

Camount

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using coin as the loop variable which conflicts with the coin value

Using amount as loop variable which is unrelated

Using index but not consistent in the loop condition

The variable i is commonly used as the index in loops to iterate over arrays.

3fill in blank

hard

Fix the error in the inner loop to update the ways array correctly.

DSA Javascript

for (let j = coin; j <= amount; j++) {
  ways[j] = ways[j] [1] ways[j - coin];
}

Drag options to blanks, or click blank then click option'

A*

B-

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction which reduces the count incorrectly

Using multiplication or division which are not correct for counting ways

We add ways[j - coin] to ways[j] because each coin can contribute to the total ways to make amount j.

4fill in blank

hard

Fill both blanks to complete the function header and return statement.

DSA Javascript

function coinChange([1], [2]) {
  // function body
  return ways[amount];
}

Drag options to blanks, or click blank then click option'

Acoins

Bamount

Ctotal

Dvalues

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect parameter names like total or values

Swapping the order of parameters

The function takes coins array and amount as parameters and returns ways[amount].

5fill in blank

hard

Fill all three blanks to complete the inner loop and update ways array correctly.

DSA Javascript

for (let [1] = [2]; [1] <= [3]; [1]++) {
  ways[j] += ways[j - coin];
}

Drag options to blanks, or click blank then click option'

Aj

Bcoin

Camount

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of j as loop variable

Starting loop at 0 instead of coin

Ending loop before amount

The loop variable is j starting from coin up to amount to update ways array.

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