Coin Change Total Number of Ways in DSA Typescript - Execution Trace

Concept Flow - Coin Change Total Number of Ways

Start with empty ways array

↓

Initialize ways[0

↓

For each coin

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Result: ways[target

We build an array where each index shows ways to make that amount. For each coin, we add ways to make amounts including that coin.

Execution Sample

DSA Typescript

const coins = [1, 2, 3];
const amount = 4;
const ways = Array(amount + 1).fill(0);
ways[0] = 1;
for (const coin of coins) {
  for (let i = coin; i <= amount; i++) {
    ways[i] += ways[i - coin];
  }
}

Calculate total ways to make amount 4 using coins 1, 2, and 3.

Execution Table

Step	Operation	Amount Index Updated	ways Array State	Explanation
0	Initialize ways	0	[1, 0, 0, 0, 0]	ways[0] = 1 means 1 way to make amount 0 (use no coins)
1	Use coin 1	1	[1, 1, 0, 0, 0]	ways[1] += ways[0] → 0 + 1 = 1
2	Use coin 1	2	[1, 1, 1, 0, 0]	ways[2] += ways[1] → 0 + 1 = 1
3	Use coin 1	3	[1, 1, 1, 1, 0]	ways[3] += ways[2] → 0 + 1 = 1
4	Use coin 1	4	[1, 1, 1, 1, 1]	ways[4] += ways[3] → 0 + 1 = 1
5	Use coin 2	2	[1, 1, 2, 1, 1]	ways[2] += ways[0] → 1 + 1 = 2
6	Use coin 2	3	[1, 1, 2, 2, 1]	ways[3] += ways[1] → 1 + 1 = 2
7	Use coin 2	4	[1, 1, 2, 2, 3]	ways[4] += ways[2] → 1 + 2 = 3
8	Use coin 3	3	[1, 1, 2, 3, 3]	ways[3] += ways[0] → 2 + 1 = 3
9	Use coin 3	4	[1, 1, 2, 3, 4]	ways[4] += ways[1] → 3 + 1 = 4
10	End	-	[1, 1, 2, 3, 4]	All coins processed, ways[4] = 4 total ways

💡 All coins processed, final ways array shows total ways to make each amount

Variable Tracker

Variable	Start	After Step 1	After Step 5	After Step 8	Final
ways	[1,0,0,0,0]	[1,1,0,0,0]	[1,1,2,1,1]	[1,1,2,3,3]	[1,1,2,3,4]
coin	N/A	1	2	3	3
i (amount index)	N/A	1	2	3	4

Key Moments - 3 Insights

Why do we set ways[0] = 1 at the start?

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This is the base case that allows building up solutions for larger amounts (see execution_table step 0).

Why do we add ways[i - coin] to ways[i] inside the loop?

Adding ways[i - coin] means we count all ways to make the smaller amount (i - coin) and then add the current coin to reach amount i. This accumulates all combinations including the current coin (see execution_table steps 1-9).

Why do we loop coins first, then amounts, not the other way?

Looping coins first ensures combinations are counted without duplicates. Each coin adds ways to amounts in order, so we don't count the same combination multiple times in different orders (see concept_flow).

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the value of ways[4]?

A 1

B 2

C 3

D 4

💡 Hint

Check the 'ways Array State' column at step 7 in execution_table

At which step does ways[3] first become 3?

A Step 8

B Step 6

C Step 3

D Step 9

💡 Hint

Look at the 'ways Array State' column for ways[3] in execution_table rows

If we remove coin 1 from the coins array, how would ways[4] change after processing all coins?

A It would become 3

B It would become 1

C It would stay 4

D It would become 0

💡 Hint

Without coin 1, fewer combinations exist; check how coin 1 updates ways in execution_table steps 1-4

Concept Snapshot

Coin Change Total Number of Ways:
- Use a ways array of size amount+1
- Initialize ways[0] = 1 (base case)
- For each coin, update ways[i] += ways[i - coin] for i >= coin
- ways[amount] gives total combinations
- Order of loops avoids duplicates

Full Transcript

This visualization shows how to find the total number of ways to make a target amount using given coin denominations. We start with an array 'ways' where ways[0] = 1, meaning one way to make zero amount. For each coin, we update the ways array by adding ways to make smaller amounts. The execution table traces each update step by step, showing how the array changes. Key moments clarify why ways[0] is set to 1, why we add ways[i - coin], and why the coin loop comes first. The quiz tests understanding of array states at specific steps and effects of removing coins. The snapshot summarizes the approach in simple steps.

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