Coin Change Total Number of Ways in DSA Javascript

Mental Model

We want to find how many different ways we can make a total amount using given coin values, counting all combinations.

Analogy: Imagine you have different types of candies and want to fill a bag with a certain number of candies. You want to know all the ways to pick candies so the total matches exactly.

coins: [1, 2, 3]
amount: 4

Ways table (rows: coins, columns: amounts):
  0 1 2 3 4
1 1 ? ? ? ?
2 1 ? ? ? ?
3 1 ? ? ? ?

'1' means there is exactly 1 way to make amount 0 (use no coins).

Dry Run Walkthrough

Input: coins: [1, 2, 3], amount: 4

Goal: Find total number of ways to make amount 4 using coins 1, 2, and 3.

Step 1: Initialize ways array with ways[0] = 1 (one way to make amount 0).

ways: [1, 0, 0, 0, 0]

Why: We start knowing there is exactly one way to make amount zero: use no coins.

Step 2: Use coin 1: update ways for amounts 1 to 4 by adding ways[amount - 1].

ways: [1, 1, 1, 1, 1]

Why: With coin 1, each amount can be made by adding one coin 1 to the ways of amount-1.

Step 3: Use coin 2: update ways for amounts 2 to 4 by adding ways[amount - 2].

ways: [1, 1, 2, 2, 3]

Why: Coin 2 adds new combinations for amounts 2, 3, and 4 based on ways to make amounts 0, 1, and 2.

Step 4: Use coin 3: update ways for amounts 3 to 4 by adding ways[amount - 3].

ways: [1, 1, 2, 3, 4]

Why: Coin 3 adds new combinations for amounts 3 and 4 based on ways to make amounts 0 and 1.

Result:

ways: [1, 1, 2, 3, 4]
Total ways to make amount 4: 4

Annotated Code

DSA Javascript

class CoinChange {
  static totalWays(coins, amount) {
    const ways = Array(amount + 1).fill(0);
    ways[0] = 1; // base case: one way to make amount 0

    for (const coin of coins) {
      for (let a = coin; a <= amount; a++) {
        ways[a] += ways[a - coin]; // add ways to make (a - coin)
      }
    }

    return ways[amount];
  }
}

// Driver code
const coins = [1, 2, 3];
const amount = 4;
const result = CoinChange.totalWays(coins, amount);
console.log(`Total ways to make amount ${amount}: ${result}`);

ways[0] = 1; // base case: one way to make amount 0

Initialize ways array with one way to make zero amount

for (const coin of coins) {

Iterate over each coin to build up ways

for (let a = coin; a <= amount; a++) {

For each amount from coin value up to target, update ways

ways[a] += ways[a - coin]; // add ways to make (a - coin)

Add ways to make amount (a - coin) to current ways[a]

return ways[amount];

Return total ways to make the target amount

Output

Total ways to make amount 4: 4

Complexity Analysis

Time: O(n * amount) because we loop through each coin and for each coin, we loop through amounts up to the target.

Space: O(amount) because we use a single array of size amount+1 to store ways counts.

vs Alternative: Compared to recursive solutions without memoization, this dynamic programming approach avoids repeated calculations and is much faster.

Edge Cases

amount is 0

Returns 1 because there is exactly one way to make amount zero (use no coins).

DSA Javascript

ways[0] = 1; // base case: one way to make amount 0

coins array is empty

Returns 0 because no coins means no way to make positive amounts.

DSA Javascript

for (const coin of coins) {

amount less than smallest coin

Returns 0 because no coin can be used to make the amount.

DSA Javascript

for (let a = coin; a <= amount; a++) {

When to Use This Pattern

When asked to count the number of ways to make a sum from given coin denominations, use the Coin Change Total Number of Ways pattern with dynamic programming to build solutions bottom-up.

Common Mistakes

Mistake: Using a nested loop that counts permutations instead of combinations, leading to overcounting.
Fix: Iterate coins in the outer loop and amounts in the inner loop to count combinations correctly.

Mistake: Not initializing ways[0] = 1, causing zero ways for amount zero and incorrect results.
Fix: Always set ways[0] = 1 before starting the loops.

Summary

Calculates how many different combinations of coins can make a target amount.

Use when you need to count all possible ways to form a sum from given coin values.

The key is to build up the number of ways incrementally by adding ways from smaller amounts.

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