Coin Change Total Number of Ways in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the 'Coin Change Total Number of Ways' problem ask you to find?

It asks for the total number of different ways to make a target amount using given coin denominations, where coins can be used unlimited times.

beginner

In the Coin Change problem, why do we use dynamic programming?

Because it helps us build the solution step-by-step by storing results of smaller amounts to avoid repeated calculations, making the process efficient.

beginner

What does the dp array represent in the Coin Change Total Number of Ways solution?

Each index in the dp array represents the number of ways to make that amount using the given coins.

intermediate

Why do we iterate over coins first, then amounts, in the Coin Change solution?

Iterating coins first ensures combinations are counted without duplicates, as each coin is considered in order to build up amounts.

beginner

What is the base case for the dp array in the Coin Change Total Number of Ways problem?

dp[0] = 1, meaning there is exactly one way to make amount zero — by choosing no coins.

What does dp[i] represent in the Coin Change Total Number of Ways problem?

A Number of ways to make amount i

B Minimum coins needed to make amount i

C Maximum coins used to make amount i

D Number of coins available

dp[i] stores the total number of ways to make amount i using the given coins.

Why is dp[0] initialized to 1 in the Coin Change problem?

A Because amount zero can be made in one way: using no coins

B Because zero coins are available

C Because zero is the smallest coin

D Because dp arrays start with 1 by default

There is exactly one way to make amount zero: by choosing no coins.

In the Coin Change problem, which loop order avoids counting duplicate combinations?

A Loop over amounts first, then coins

B Loop over coins first, then amounts

C Loop over coins twice

D Loop over amounts twice

Looping over coins first ensures combinations are counted uniquely without duplicates.

If coins = [1, 2] and amount = 3, how many ways are there to make 3?

A 3

B 4

C 1

D 2

Ways: (1+1+1), (1+2). Since order does not matter, total unique combinations are 2 ways: (1+1+1) and (1+2).

What is the time complexity of the Coin Change Total Number of Ways dynamic programming solution?

A O(n^2)

B O(amount^2)

C O(n * amount), where n is number of coins

D O(n + amount)

The solution uses nested loops over coins and amounts, resulting in O(n * amount) time.

Explain how to use dynamic programming to find the total number of ways to make a target amount with given coins.

Think about building solutions from smaller amounts to bigger amounts.

Describe why the order of loops matters in the Coin Change Total Number of Ways problem.

Consider how counting changes if you swap loops.

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