Coin Change Total Number of Ways in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of the following JavaScript code that calculates the total number of ways to make change for amount 5 using coins [1, 2, 5]?

DSA Javascript

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

console.log(coinChangeWays([1, 2, 5], 5));

A 5

B 4

C 3

D 6

💡 Hint

Think about how many combinations of coins can sum to 5 using 1, 2, and 5.

The code uses dynamic programming to count ways. For amount 5 with coins [1,2,5], the ways are: {1+1+1+1+1}, {1+1+1+2}, {1+2+2}, {5}, totaling 4 ways.

❓ Predict Output

intermediate

Result of Coin Change with Different Coins

What does this code print for amount 4 with coins [1, 3, 4]?

DSA Javascript

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

console.log(coinChangeWays([1, 3, 4], 4));

A 3

B 1

C 4

D 2

💡 Hint

Count all combinations of coins that sum to 4.

Ways to make 4: {1+1+1+1}, {1+3}, {4} → 3 ways total.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does this code produce when run, and why?

DSA Javascript

function coinChangeWays(coins, amount) {
  const dp = Array(amount + 1).fill(0);
  dp[0] = 1;
  for (let i = 0; i <= coins.length; i++) {
    for (let j = coins[i]; j <= amount; j++) {
      dp[j] += dp[j - coins[i]];
    }
  }
  return dp[amount];
}

console.log(coinChangeWays([1, 2, 3], 4));

A ReferenceError: coins is not defined

B RangeError: Maximum call stack size exceeded

C TypeError: Cannot read property 'length' of undefined

D TypeError: Cannot read property '0' of undefined

💡 Hint

Check the loop condition for accessing coins array.

The loop runs with i <= coins.length, so when i equals coins.length, coins[i] is undefined causing dp[j - coins[i]] to fail with TypeError.

🧠 Conceptual

advanced

Understanding DP Array in Coin Change

In the coin change total ways algorithm, what does dp[i] represent after the loops complete?

A The maximum number of coins that can be used to make amount i

B The minimum number of coins needed to make amount i

C The number of ways to make change for amount i using all coins considered so far

D The total sum of coin values up to amount i

💡 Hint

Think about what the dp array accumulates as the algorithm progresses.

dp[i] stores the count of different ways to make amount i using the coins processed up to that point.

❓ Predict Output

expert

Output of Coin Change with Large Amount and Coins

What is the output of this code for amount 10 with coins [2, 5, 3, 6]?

DSA Javascript

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

console.log(coinChangeWays([2, 5, 3, 6], 10));

A 6

B 5

C 4

D 7

💡 Hint

Count all combinations of coins that sum to 10.

The ways to make 10 with coins [2,3,5,6] are 6 distinct combinations.

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