Coin Change Total Number of Ways in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of the following Java code that calculates the total number of ways to make change for amount 5 using coins {1, 2, 3}?

DSA Java

public class CoinChange {
    public static int countWays(int[] coins, int amount) {
        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];
            }
        }
        return dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 2, 3};
        int amount = 5;
        System.out.println(countWays(coins, amount));
    }
}

A 7

B 6

C 5

D 8

💡 Hint

Think about how many combinations of coins can sum up to 5 using 1, 2, and 3.

The code uses dynamic programming to count ways. For amount 5 with coins {1,2,3}, the total ways are 7.

❓ Predict Output

intermediate

Output of Coin Change Ways with Different Coins

What is the output of this Java code that calculates the total number of ways to make change for amount 4 using coins {1, 3, 4}?

DSA Java

public class CoinChange {
    public static int countWays(int[] coins, int amount) {
        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];
            }
        }
        return dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 3, 4};
        int amount = 4;
        System.out.println(countWays(coins, amount));
    }
}

A 5

B 3

C 2

D 4

💡 Hint

Count all combinations of coins that sum to 4.

The code counts ways to make 4 with coins {1,3,4}. The ways are: {1,1,1,1}, {1,3}, {4}, total 3 ways.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error will this Java code produce when trying to calculate the number of ways to make change for amount 5 with coins {1, 2}?

DSA Java

public class CoinChange {
    public static int countWays(int[] coins, int amount) {
        int[] dp = new int[amount];
        dp[0] = 1;
        for (int coin : coins) {
            for (int i = coin; i <= amount; i++) {
                dp[i] += dp[i - coin];
            }
        }
        return dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 2};
        int amount = 5;
        System.out.println(countWays(coins, amount));
    }
}

A NullPointerException

B No error, outputs 5

C ArithmeticException

D ArrayIndexOutOfBoundsException

💡 Hint

Check the size of the dp array and the indices accessed.

The dp array is of size amount (5), but code accesses dp[amount] which is dp[5], out of bounds since indices go 0 to 4.

🧠 Conceptual

advanced

Understanding DP Array Initialization in Coin Change

Why is dp[0] initialized to 1 in the coin change total number of ways dynamic programming solution?

A Because dp[0] stores the total number of coins available

B Because dp[0] is used to store the maximum coin value

C Because there is exactly one way to make amount 0 — by choosing no coins

D Because dp[0] counts the number of coin types

💡 Hint

Think about how many ways you can make zero amount.

dp[0] = 1 means one way to make amount zero: use no coins at all.

🚀 Application

expert

Number of Ways to Make Change for Large Amount

Given coins {1, 5, 10, 25} and amount 30, what is the total number of ways to make change? Use the standard coin change total ways approach.

A 22

B 14

C 20

D 18

💡 Hint

Use dynamic programming or think about combinations of coins that sum to 30.

The total ways to make 30 with coins {1,5,10,25} is 22 using the classic DP approach.

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