Coin Change Minimum Coins in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with a large number.

DSA Java

int[] dp = new int[amount + 1];
for (int i = 1; i <= amount; i++) {
    dp[i] = [1];
}

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

Aamount

B0

CInteger.MAX_VALUE

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 will make the dp array incorrect because 0 means no coins needed.

Using -1 can cause errors in calculations.

Using amount is not large enough to represent unreachable.

We use Integer.MAX_VALUE to represent an initially unreachable state for each amount.

2fill in blank

medium

Complete the code to set the base case for amount 0.

DSA Java

dp[0] = [1];

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

A0

B1

C-1

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to -1 or 1 causes wrong results.

Using amount is incorrect for base case.

No coins are needed to make amount 0, so dp[0] = 0.

3fill in blank

hard

Fix the error in the condition to check if the coin can be used.

DSA Java

for (int coin : coins) {
    if (coin [1] i) {
        dp[i] = Math.min(dp[i], dp[i - coin] + 1);
    }
}

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

A<

B>

C>=

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or >= will skip valid coins.

Using < excludes coins equal to the amount.

We check if the coin value is less than or equal to the current amount i to use it.

4fill in blank

hard

Fill both blanks to return the result or -1 if no solution.

DSA Java

return dp[amount] == [1] ? [2] : dp[amount];

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

AInteger.MAX_VALUE

B-1

C0

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 for no solution.

Comparing dp[amount] to 0 instead of Integer.MAX_VALUE.

If dp[amount] is still Integer.MAX_VALUE, it means no solution was found, so return -1.

5fill in blank

hard

Fill all three blanks to complete the main loop updating dp array.

DSA Java

for (int i = [1]; i <= amount; i++) {
    for (int coin : coins) {
        if (coin [2] i && dp[i - coin] != [3]) {
            dp[i] = Math.min(dp[i], dp[i - coin] + 1);
        }
    }
}

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

A1

B<=

CInteger.MAX_VALUE

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 causes redundant checks.

Using < instead of <= excludes valid coins.

Not checking dp[i - coin] != Integer.MAX_VALUE causes errors.

Start from 1 to amount, check coin <= i, and ensure dp[i - coin] is reachable (not Integer.MAX_VALUE).

Practice

1. What does the Coin Change Minimum Coins problem ask you to find?
Given coins of different denominations and a total amount, find the minimum number of coins needed to make that amount.

easy

A. Minimum number of coins needed to make the amount

B. Maximum number of coins needed to make the amount

C. Number of ways to make the amount

D. Sum of all coin denominations

Solution

1. Step 1: Understand the problem statement

   The problem asks for the least number of coins needed to reach a given amount using available coin denominations.

2. Step 2: Differentiate from similar problems

   It is not about counting ways or maximum coins, but the minimum coins required.

3. Final Answer:

   Minimum number of coins needed to make the amount -> Option A

4. Quick Check:

   Minimum coins = minimum coins needed [OK]

Hint: Focus on 'minimum coins' needed, not ways or sums [OK]

Common Mistakes:

• Confusing minimum coins with number of ways
• Thinking about maximum coins instead of minimum
• Ignoring the total amount requirement

2. Which of the following Java method signatures correctly represents a function to find minimum coins for a given amount?

public static int minCoins(int[] coins, int amount)

easy

A. public int minCoins(int coins, int[] amount)

B. public static int minCoins(int[] coins, int amount)

C. static void minCoins(int[] coins, int amount)

D. public static int minCoins(int amount, int[] coins)

Solution

1. Step 1: Check parameter types and order

   The method should accept an array of coin denominations and an integer amount, in that order.

2. Step 2: Check return type and modifiers

   It should return an int representing minimum coins and be static for utility use.

3. Final Answer:

   public static int minCoins(int[] coins, int amount) -> Option B

4. Quick Check:

   Correct signature = public static int minCoins(int[] coins, int amount) [OK]

Hint: Array first, amount second; return int minimum coins [OK]

Common Mistakes:

• Swapping parameter order
• Wrong return type (void instead of int)
• Missing static keyword if utility method

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

public class CoinChange {
    public static int minCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        java.util.Arrays.fill(dp, max);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }

    public static void main(String[] args) {
        int[] coins = {1, 3, 4};
        int amount = 6;
        System.out.println(minCoins(coins, amount));
    }
}

medium

A. -1

B. 3

C. 1

D. 2

Solution

1. Step 1: Understand the DP array filling

   dp[i] stores minimum coins for amount i. Initialized with max (7 here) except dp[0]=0.

2. Step 2: Calculate dp values for amount=6

   Using coins {1,3,4}: - dp[1] = 1 (1 coin of 1) - dp[2] = 2 (1+1) - dp[3] = 1 (coin 3) - dp[4] = 1 (coin 4) - dp[5] = 2 (4+1) - dp[6] = min(dp[6-1]+1=dp[5]+1=3, dp[6-3]+1=dp[3]+1=2, dp[6-4]+1=dp[2]+1=3) = 2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array updates for each amount [OK]

Common Mistakes:

• Not initializing dp array correctly
• Returning dp[amount] without checking if solution exists
• Confusing dp indices or coin subtraction

4. Find the error in the following Java code snippet for minimum coin change:

public static int minCoins(int[] coins, int amount) {
    int[] dp = new int[amount + 1];
    Arrays.fill(dp, Integer.MAX_VALUE);
    dp[0] = 0;
    for (int i = 1; i <= amount; i++) {
        for (int coin : coins) {
            if (coin <= i) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
    }
    return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}

medium

A. Using Integer.MAX_VALUE causes overflow when adding 1

B. dp array size should be amount, not amount + 1

C. dp[0] should be Integer.MAX_VALUE, not 0

D. The for loop should start from 0, not 1

Solution

1. Step 1: Identify the problem with Integer.MAX_VALUE

   Adding 1 to Integer.MAX_VALUE causes integer overflow, leading to incorrect dp values.

2. Step 2: Explain correct initialization

   Use a large number less than Integer.MAX_VALUE (like amount+1) to avoid overflow when adding 1.

3. Final Answer:

   Using Integer.MAX_VALUE causes overflow when adding 1 -> Option A

4. Quick Check:

   Integer.MAX_VALUE + 1 overflows [OK]

Hint: Avoid Integer.MAX_VALUE for dp initialization to prevent overflow [OK]

Common Mistakes:

• Ignoring overflow when adding 1 to max value
• Wrong dp array size
• Incorrect dp[0] initialization

5. Given coin denominations {2, 5, 10} and amount 7, what is the minimum number of coins needed? If not possible, return -1.
Use the Coin Change Minimum Coins approach.

medium

A. 1

B. 3

C. 2

D. -1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins: 2, 5, 10 - 2+5=7 (2 coins) - 5+2=7 (2 coins) - 10 is too big But 2+5=7 is valid, so minimum coins = 2.

2. Step 2: Verify if amount 7 can be formed

   Actually, 2+5=7 is valid, so minimum coins = 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   7 = 5 + 2 = 2 coins [OK]

Hint: Try largest coin first, then smaller coins [OK]

Common Mistakes:

• Assuming no solution without checking combinations
• Ignoring order of coins
• Returning -1 prematurely