Coin Change Total Number of Ways in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array for coin change.

DSA Java

int[] dp = new int[[1] + 1];

Drag options to blanks, or click blank then click option'

Acoins.length

Bcoins.length + 1

Camount + 1

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Using coins.length instead of amount for dp size.

Forgetting to add 1 to amount for dp array size.

The dp array size should be amount + 1 to represent all amounts from 0 to amount.

2fill in blank

medium

Complete the code to set the base case for dp array.

DSA Java

dp[[1]] = 1;

Drag options to blanks, or click blank then click option'

A1

B0

Camount

Dcoins.length

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[1] = 1 instead of dp[0].

Setting dp[amount] = 1 incorrectly.

There is exactly one way to make amount 0: use no coins.

3fill in blank

hard

Fix the error in the inner loop condition to avoid index out of bounds.

DSA Java

for (int j = [1]; j <= amount; j++) {

Drag options to blanks, or click blank then click option'

Acoins[i]

Bcoins[i] + 1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting j from 0 causes negative index access.

Starting j from 1 causes errors if coin value > 1.

The inner loop starts from coins[i] to ensure valid indices when accessing dp[j - coins[i]].

4fill in blank

hard

Fill both blanks to update dp array correctly inside nested loops.

DSA Java

dp[j] = dp[j] [1] dp[j [2] coins[i]];

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Using addition instead of subtraction inside index.

We add the ways to make amount j without current coin and ways to make amount j - coin value.

5fill in blank

hard

Fill all three blanks to complete the coin change function returning total ways.

DSA Java

public int change(int [1], int[] [2]) {
    int[] dp = new int[[3] + 1];
    dp[0] = 1;
    for (int i = 0; i < coins.length; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}

Drag options to blanks, or click blank then click option'

Aamount

Bcoins

Dcoin

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter names.

Incorrect dp array size.

The function parameters are amount and coins array; dp array size is amount + 1.

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