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

Concept Flow - Coin Change Total Number of Ways

Start with amount = 0

↓

Initialize ways array with ways[0

↓

For each coin in coins

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Result is ways[target

We build up the number of ways to make each amount by adding ways contributed by each coin step-by-step.

Execution Sample

DSA Javascript

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 x = coin; x <= amount; x++) {
    ways[x] += ways[x - coin];
  }
}
console.log(ways[amount]);

This code calculates how many ways to make amount 4 using coins 1, 2, and 3.

Execution Table

Step	Operation	Amount Index (x)	ways[x] Before	ways[x - coin]	ways[x] After	Visual ways Array
Init	Set ways[0] = 1	0	0	-	1	[1, 0, 0, 0, 0]
Coin=1	x=1	1	0	ways[0]=1	1	[1, 1, 0, 0, 0]
Coin=1	x=2	2	0	ways[1]=1	1	[1, 1, 1, 0, 0]
Coin=1	x=3	3	0	ways[2]=1	1	[1, 1, 1, 1, 0]
Coin=1	x=4	4	0	ways[3]=1	1	[1, 1, 1, 1, 1]
Coin=2	x=2	2	1	ways[0]=1	2	[1, 1, 2, 1, 1]
Coin=2	x=3	3	1	ways[1]=1	2	[1, 1, 2, 2, 1]
Coin=2	x=4	4	1	ways[2]=2	3	[1, 1, 2, 2, 3]
Coin=3	x=3	3	2	ways[0]=1	3	[1, 1, 2, 3, 3]
Coin=3	x=4	4	3	ways[1]=1	4	[1, 1, 2, 3, 4]
End	All coins processed	-	-	-	-	[1, 1, 2, 3, 4]

💡 All coins processed, ways[4] = 4 means 4 ways to make amount 4

Variable Tracker

Variable	Start	After Coin=1	After Coin=2	After Coin=3	Final
ways	[1,0,0,0,0]	[1,1,1,1,1]	[1,1,2,2,3]	[1,1,2,3,4]	[1,1,2,3,4]
coin	-	1	2	3	3
x	-	1 to 4	2 to 4	3 to 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 ways for higher amounts (see execution_table Init row).

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

ways[x - coin] represents the number of ways to make the smaller amount before adding the current coin. Adding it to ways[x] accumulates all ways including the current coin (see rows where ways[x] changes).

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

Looping coins first ensures combinations are counted without duplicates. If we looped amounts first, order would matter and count permutations instead (see concept_flow).

Visual Quiz - 3 Questions

Look at the execution_table at step 'Coin=2, x=4'. What is ways[4] after this step?

A 3

B 2

C 4

D 1

💡 Hint

Check the 'ways[x] After' column for step 'Coin=2, x=4' in execution_table.

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

A Coin=2, x=3

B Coin=3, x=3

C Coin=1, x=3

D Coin=3, x=4

💡 Hint

Look at the 'ways[x] After' for ways[3] in execution_table rows.

If we remove coin 1 from coins, what will ways[4] be at the end?

A 2

B 3

C 1

D 4

💡 Hint

Without coin 1, ways array updates start from coin 2; check how many ways to make 4 with coins 2 and 3.

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[x] += ways[x - coin] for x >= coin
- Result is ways[amount]
- Counts combinations, not permutations

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 ways[0] = 1 because there is one way to make zero amount: no coins. Then for each coin, we update the ways array by adding the number of ways to make smaller amounts. The execution table traces each update step-by-step, showing how the ways array changes. Key points include why ways[0] is set to 1, why we add ways[x - coin] to ways[x], and why the outer loop is over coins to avoid counting duplicates. The final answer is ways[amount], which in the example is 4 ways to make amount 4 with coins 1, 2, and 3.

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