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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with zeros.

DSA Typescript

const dp: number[] = new Array([1]).fill(0);

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

Acoins.length

Bamount + 1

Camount

Dcoins.length + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using coins.length instead of amount + 1 for dp size.

Forgetting to add 1 to amount for dp array length.

We need dp array size to be amount + 1 to cover all amounts from 0 to amount.

2fill in blank

medium

Complete the code to set the base case for dp.

DSA Typescript

dp[0] = [1];

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

Aamount

B0

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to 0 which means no ways to make zero amount.

Setting dp[0] to amount which is incorrect.

There is exactly one way to make amount 0: use no coins.

3fill in blank

hard

Fix the error in the inner loop condition to avoid out-of-bound errors.

DSA Typescript

for (let j = coin; j [1] dp.length; j++) {

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of bounds error.

Using '>' or '>=' causes the loop to never run or run incorrectly.

The loop should run while j is less than dp.length to avoid index out of bounds.

4fill in blank

hard

Fill both blanks to correctly update dp array inside the loop.

DSA Typescript

dp[j] = dp[j] [1] dp[j [2] coin];

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Using j + coin instead of j - coin inside dp index.

We add the ways to make amount j and ways to make amount j - coin.

5fill in blank

hard

Fill all three blanks to complete the function that returns total ways to make amount.

DSA Typescript

function change(amount: number, coins: number[]): number {
  const dp: number[] = new Array([1]).fill(0);
  dp[0] = [2];
  for (const coin of coins) {
    for (let j = coin; j [3] dp.length; j++) {
      dp[j] = dp[j] + dp[j - coin];
    }
  }
  return dp[amount];
}

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

Aamount + 1

B1

C<

Dcoins.length

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong dp array size.

Incorrect base case value.

Wrong loop condition causing errors.

dp array size is amount + 1, base case dp[0] = 1, loop runs while j < dp.length.

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 sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible ways to make a target amount using given coins.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, not minimum count.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

4. Quick Check:

   Counting ways = total combinations [OK]

Hint: Count combinations, not minimum coins needed [OK]

Common Mistakes:

• Confusing with minimum coin count problem
• Thinking about maximum coin value instead of ways
• Ignoring repeated use of coins

2. Which of the following TypeScript code snippets correctly initializes the DP array for the coin change ways problem for amount 5?

easy

A. const dp = new Array(5).fill(1);

B. const dp = new Array(6).fill(0); dp[0] = 1;

C. const dp = new Array(5).fill(0); dp[1] = 1;

D. const dp = new Array(6).fill(1); dp[0] = 0;

Solution

1. Step 1: Understand DP array size and initialization

   For amount 5, dp array size should be 6 (0 to 5). dp[0] = 1 means one way to make amount 0.

2. Step 2: Check each option

   const dp = new Array(6).fill(0); dp[0] = 1; correctly creates dp of size 6, fills 0, sets dp[0]=1. Others have wrong size or initialization.

3. Final Answer:

   const dp = new Array(6).fill(0); dp[0] = 1; -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Using dp array size equal to amount instead of amount+1
• Not setting dp[0] = 1
• Filling dp with 1 instead of 0 initially

3. What is the output of the following TypeScript 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. 3

B. 0

C. 1

D. 2

Solution

1. Step 1: Trace dp updates for coin 1

   dp after coin 1: dp[1]=1, dp[2]=1, dp[3]=1 (only using coin 1)

2. Step 2: Trace dp updates for coin 2

   dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; final dp = [1,1,2,2]

3. Final Answer:

   2 -> Option D

4. Quick Check:

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

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

Common Mistakes:

• Not updating dp correctly inside nested loops
• Confusing dp[x] with dp[x - coin]
• Missing dp[0] = 1 initialization

4. Find the error in this TypeScript code for coin change ways and select the fix:

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

medium

A. Use for (const coin of coins) instead of index loop

B. Initialize dp array with 1 instead of 0

C. Change loop to for (let i = 0; i < coins.length; i++)

D. Set dp[amount] = 0 before loops

Solution

1. Step 1: Identify loop index error

   The loop runs i from 0 to coins.length inclusive, causing out-of-bounds access at coins[coins.length].

2. Step 2: Fix loop condition

   Change to i < coins.length to avoid accessing undefined coin.

3. Final Answer:

   Change loop to for (let i = 0; i < coins.length; i++) -> Option C

4. Quick Check:

   Loop index must be less than array length [OK]

Hint: Use < length, not <= length for array loops [OK]

Common Mistakes:

• Using <= in for loop causing out-of-bounds error
• Not initializing dp[0] = 1
• Confusing index and value in loops

5. Given coins = [1, 3, 5] and amount = 7, how many total ways are there to make amount 7 using any number of these coins?

hard

A. 4

B. 8

C. 7

D. 6

Solution

1. Step 1: Initialize dp array and process coin 1

   dp size 8, dp[0]=1. Using coin 1, dp[x]=1 for x=1 to 7.

2. Step 2: Update dp with coin 3

   For x=3 to 7, dp[x]+=dp[x-3]: dp[3]=2, dp[4]=1+dp[1]=2, dp[5]=1+dp[2]=2, dp[6]=1+dp[3]=3, dp[7]=1+dp[4]=3.

3. Step 3: Update dp with coin 5

   For x=5 to 7, dp[5]+=dp[0]=2+1=3, dp[6]+=dp[1]=3+1=4, dp[7]+=dp[2]=3+1=4.

4. Step 4: Final dp array and answer

   Final dp[7] = 4 ways. The combinations are: 7x1, 4x1+1x3, 1x1+2x3, 2x1+1x5.

5. Final Answer:

   4 -> Option A

6. Quick Check:

   Count all combinations with coins [1,3,5] for 7 = 4 [OK]

Hint: Build dp adding ways coin by coin [OK]

Common Mistakes:

• Missing some combinations with multiple coins
• Confusing permutations with combinations
• Incorrect dp updates order