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

Concept Flow - Coin Change Total Number of Ways

Start with amount = 0

↓

Initialize ways array with ways[0

↓

For each coin

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Result is ways[target amount

We build up the number of ways to make each amount by adding ways contributed by each coin, starting from 0 up to the target amount.

Execution Sample

DSA Java

int[] coins = {1, 2, 3};
int amount = 4;
int[] ways = new int[amount + 1];
ways[0] = 1;
for (int coin : coins) {
  for (int i = coin; i <= amount; i++) ways[i] += ways[i - coin];
}

This code calculates the total number of ways to make amount 4 using coins 1, 2, and 3.

Execution Table

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

💡 All coins processed and amounts updated, final ways array shows total ways to make each amount.

Variable Tracker

Variable	Start	After Step 5	After Step 8	After Step 10	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
i (amount index)	-	4	4	4	-

Key Moments - 3 Insights

Why do we start ways[0] with 1 instead of 0?

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 solutions for larger amounts (see Step 1 in execution_table).

Why do we add ways[i - coin] to ways[i] instead of replacing it?

We add because ways[i] accumulates all possible ways from different coins. Each coin contributes additional ways without removing previous ones (see Steps 6-10 where ways[i] increases).

Why does the inner loop start from coin value instead of 0?

Amounts less than the coin value cannot use that coin, so we start from coin to avoid invalid indices and unnecessary calculations (see Step 2 where i starts at 1 for coin=1).

Visual Quiz - 3 Questions

Look at the execution_table at Step 6, what is the value of ways[2] after using coin 2?

A 1

B 2

C 0

D 3

💡 Hint

Check the 'ways Array State' column at Step 6 in execution_table.

At which step does ways[4] first become greater than 1?

A Step 10

B Step 5

C Step 8

D Step 3

💡 Hint

Look at the 'ways Array State' for ways[4] in execution_table rows.

If we remove coin 1 from the coins array, what would be the value of ways[1] after processing all coins?

A 0

B 1

C 2

D 3

💡 Hint

Without coin 1, ways[1] cannot be formed; check variable_tracker for ways array start and updates.

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 ways to make amount
- Builds solutions bottom-up by adding coin contributions

Full Transcript

The Coin Change Total Number of Ways problem counts how many ways to make a target amount using given coin denominations. We use a ways array where ways[i] stores the number of ways to make amount i. We start with ways[0] = 1 because there is one way to make zero amount: using no coins. Then for each coin, we update the ways array by adding ways[i - coin] to ways[i], meaning we add the ways to make the smaller amount i - coin to the current amount i. This process accumulates all possible combinations. The final answer is ways[amount]. The execution table shows step-by-step how the ways array changes as each coin is processed. Key points include why ways[0] starts at 1, why we add instead of replace, and why the inner loop starts at the coin value. The visual quiz tests understanding of these updates and the final ways array values.

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 given amount using unlimited coins of given denominations

B. The minimum number of coins needed to make a given amount

C. The maximum value of coins that can be used to make a given amount

D. The total number of coins available in the input array

Solution

1. Step 1: Understand the problem statement

   The problem asks for counting all possible ways to form a target amount using unlimited coins of given denominations.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, here we count ways, not minimize coins.

3. Final Answer:

   The total number of ways to make a given amount using unlimited coins of given denominations -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

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

Common Mistakes:

• Confusing with minimum coins problem
• Thinking about maximum coin value
• Counting coins instead of ways

2. Which of the following Java code snippets correctly initializes the DP array for the Coin Change Total Number of Ways problem for amount 5?

easy

A. int[] dp = new int[6]; dp[0] = 1;

B. int[] dp = new int[5]; dp[0] = 1;

C. int[] dp = new int[6]; dp[1] = 1;

D. int[] dp = new int[5]; dp[1] = 1;

Solution

1. Step 1: Determine DP array size

   For amount 5, dp array size should be 6 to include index 0 to 5.

2. Step 2: Initialize base case

   dp[0] = 1 means one way to make amount 0 (using no coins).

3. Final Answer:

   int[] dp = new int[6]; 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 dp array size equal to amount instead of amount+1
• Initializing dp[1] instead of dp[0]
• Forgetting to set base case dp[0] = 1

3. What is the output of the following Java code snippet?

int[] coins = {1, 2};
int amount = 3;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
  for (int i = coin; i <= amount; i++) {
    dp[i] += dp[i - coin];
  }
}
System.out.println(dp[amount]);

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] initially, dp[0] = 1.

2. Step 2: Process coin 1

   For i=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

   For i=2 to 3: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; dp = [1,1,2,2]

4. Step 4: Final dp[3]

   dp[3] = 2 means 2 ways: {1+1+1, 1+2}

5. Final Answer:

   2 -> Option C

6. Quick Check:

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

Hint: Add ways for each coin from coin to amount [OK]

Common Mistakes:

• Not updating dp correctly inside nested loops
• Confusing dp[i] with dp[i-1]
• Miscounting combinations

4. Identify the error in the following Java code for Coin Change Total Number of Ways:

int[] coins = {1, 2, 3};
int amount = 4;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int i = 1; i <= amount; i++) {
  for (int coin : coins) {
    if (i >= coin) {
      dp[i] += dp[i - coin];
    }
  }
}
System.out.println(dp[amount]);

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition i >= coin is incorrect

D. The outer loop should iterate over coins, inner over amounts

Solution

1. Step 1: Analyze loop order

   Outer loop iterates over amounts, inner over coins, which counts permutations, not combinations.

2. Step 2: Correct loop order

   Outer loop should iterate over coins, inner over amounts to count combinations correctly.

3. Final Answer:

   The outer loop should iterate over coins, inner over amounts -> Option D

4. Quick Check:

   Loop order affects counting ways [OK]

Hint: Outer loop coins, inner loop amounts for combinations [OK]

Common Mistakes:

• Swapping loop order causing wrong counts
• Incorrect dp initialization
• Misusing conditions inside loops

5. Given coins = {1, 3, 5} and amount = 7, how many total ways are there to make amount 7 using unlimited coins? Use the Coin Change Total Number of Ways approach.

hard

A. 5

B. 4

C. 6

D. 7

Solution

1. Step 1: Initialize dp array

   dp size = 8, dp[0] = 1, others 0.

2. Step 2: Process coin 1

   Update dp for i=1 to 7: dp[i] += dp[i-1], dp = [1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   For i=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3; dp = [1,1,1,2,2,2,3,3]

4. Step 4: Process coin 5

   For i=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4; dp = [1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option B

6. Quick Check:

   Count ways with coins 1,3,5 for amount 7 = 4 [OK]

Hint: Add ways coin by coin updating dp array [OK]

Common Mistakes:

• Miscounting dp updates for each coin
• Forgetting base case dp[0] = 1
• Mixing permutations with combinations