Coin Change Minimum Coins in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

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

C-1

Dint.MaxValue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of a large number causes incorrect minimum calculations.

Using negative numbers can cause logic errors.

We initialize dp[i] with int.MaxValue to represent that the amount i is initially unreachable.

2fill in blank

medium

Complete the code to update dp[i] with the minimum coins needed.

DSA C#

for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (i >= coin && dp[i - coin] != int.MaxValue) {
            dp[i] = Math.[1](dp[i], dp[i - coin] + 1);
        }
    }
}

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

AMin

BMax

CAbs

DRound

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.Max will give the wrong result.

Using Math.Abs or Math.Round is not relevant here.

We use Math.Min to keep the smallest number of coins needed for amount i.

3fill in blank

hard

Fix the error in the return statement to handle unreachable amounts.

DSA C#

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

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

Aint.MaxValue

B0

Camount

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with 0 or -1 causes incorrect results.

Returning dp[amount] without check may return a large number.

If dp[amount] is int.MaxValue, it means the amount cannot be formed, so return -1.

4fill in blank

hard

Fill both blanks to complete the loop that iterates over amounts and coins.

DSA C#

for (int [1] = 1; [1] <= [2]; [1]++) {
    foreach (int coin in coins) {
        if ([1] >= coin && dp[[1] - coin] != int.MaxValue) {
            dp[[1]] = Math.Min(dp[[1]], dp[[1] - coin] + 1);
        }
    }
}

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

Ai

Bj

Camount

Dcoins.Length

Attempts:

3 left

💡 Hint

Common Mistakes

Using j as loop variable is confusing here.

Using coins.Length as loop limit is incorrect.

The outer loop variable is i, iterating from 1 to amount.

5fill in blank

hard

Fill all three blanks to complete the code that populates the dictionary mapping coins to their counts if count is positive.

DSA C#

var result = new Dictionary<int, int>();
for (int [3] = 0; [3] < coins.Length; [3]++) {
    if (counts[[3]] > 0) {
        result[[1]] = [2];
    }
}

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

Acoins[{{BLANK_3}}]

Bcounts[{{BLANK_3}}]

Ci

Dcoin

Attempts:

3 left

💡 Hint

Common Mistakes

Using undefined variables like coin instead of i.

Mixing up keys and values in the dictionary.

We map each coin (coins[i]) to its count (counts[i]) using index i.

Practice

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

easy

A. The smallest number of coins needed to make the amount

B. The total value of all coins

C. The maximum number of coins that can be used

D. The number of different coin types

Solution

1. Step 1: Understand the problem goal

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

2. Step 2: Clarify what is asked

   We want the smallest count of coins that sum exactly to the amount, not total value or max coins.

3. Final Answer:

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

4. Quick Check:

   Minimum coins = smallest count [OK]

Hint: Focus on minimum count, not total value or types [OK]

Common Mistakes:

• Confusing minimum coins with total coin value
• Thinking about maximum coins instead of minimum
• Counting coin types instead of coin quantity

2. Which of the following C# code snippets correctly initializes an array to store minimum coins for amounts 0 to 5, setting 0 coins for amount 0 and a large number for others?

easy

A. int[] minCoins = new int[5] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue};

B. int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue};

C. int[] minCoins = new int[6]; for(int i=1; i<=5; i++) minCoins[i] = 0;

D. int[] minCoins = new int[6]; minCoins[0] = int.MaxValue;

Solution

1. Step 1: Check array size and initialization

   We need an array of size 6 for amounts 0 to 5. Amount 0 requires 0 coins, others start with a large number.

2. Step 2: Verify each option

   int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue}; correctly initializes size 6 with 0 at index 0 and int.MaxValue for others. Others either have wrong size or wrong initialization.

3. Final Answer:

   int[] minCoins = new int[6] {0, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue, int.MaxValue}; -> Option B

4. Quick Check:

   Array size 6, 0 at start, max values after [OK]

Hint: Array size = amount + 1, 0 coins for amount 0 [OK]

Common Mistakes:

• Using wrong array size
• Setting 0 coins for all amounts
• Setting int.MaxValue for amount 0

3. What is the output of the following C# code?

int[] coins = {1, 3, 4};
int amount = 6;
int[] minCoins = new int[amount + 1];
for (int i = 1; i <= amount; i++) minCoins[i] = int.MaxValue;

for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin <= i && minCoins[i - coin] != int.MaxValue) {
            minCoins[i] = Math.Min(minCoins[i], minCoins[i - coin] + 1);
        }
    }
}
Console.WriteLine(minCoins[amount]);

medium

A. 3

B. int.MaxValue

C. 4

D. 2

Solution

1. Step 1: Initialize minCoins array

   minCoins[0] = 0 (default), others set to int.MaxValue initially.

2. Step 2: Calculate minCoins for amounts 1 to 6

   For amount 6, check coins 1,3,4:
   - Using coin 1: minCoins[5] + 1
   - Using coin 3: minCoins[3] + 1
   - Using coin 4: minCoins[2] + 1
   Calculations:
   minCoins[1] = 1 (coin 1)
   minCoins[2] = 2 (1+1)
   minCoins[3] = 1 (coin 3)
   minCoins[4] = 1 (coin 4)
   minCoins[5] = 2 (3+1 or 4+1)
   minCoins[6] = min(minCoins[5]+1, minCoins[3]+1, minCoins[2]+1) = min(3, 2, 3) = 2

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Minimum coins for 6 = 2 (3+3 or 4+1+1) [OK]

Hint: Trace minCoins array updates stepwise [OK]

Common Mistakes:

• Not updating minCoins correctly
• Ignoring int.MaxValue checks
• Miscounting coin combinations

4. Identify the error in this C# code snippet for minimum coin change and select the fix:

int[] coins = {1, 2, 5};
int amount = 11;
int[] minCoins = new int[amount + 1];
for (int i = 1; i <= amount; i++) minCoins[i] = int.MaxValue;

for (int i = 1; i <= amount; i++) {
    foreach (int coin in coins) {
        if (coin < i && minCoins[i - coin] != int.MaxValue) {
            minCoins[i] = Math.Min(minCoins[i], minCoins[i - coin] + 1);
        }
    }
}
Console.WriteLine(minCoins[amount]);

medium

A. Change loop to start from i = 0 instead of 1

B. Initialize minCoins array with zeros instead of int.MaxValue

C. Change 'coin < i' to 'coin <= i' in the if condition

D. Remove the if condition inside the foreach loop

Solution

1. Step 1: Analyze the if condition

   The condition 'coin < i' excludes cases where coin equals i, missing direct matches.

2. Step 2: Correct the condition

   Changing to 'coin <= i' allows using coins that exactly match the current amount, fixing the logic.

3. Final Answer:

   Change 'coin < i' to 'coin <= i' in the if condition -> Option C

4. Quick Check:

   Include coin equal to amount for correct min coins [OK]

Hint: Use <= to include exact coin matches [OK]

Common Mistakes:

• Using < instead of <= excludes valid coins
• Initializing minCoins incorrectly
• Removing necessary condition checks

5. You have coin denominations {2, 5, 10} and want to find the minimum coins for amount 7. Which approach correctly solves this using dynamic programming in C#?

hard

A. Initialize minCoins with 0 for amount 0 and int.MaxValue for others; update minCoins[i] by checking coins <= i and minCoins[i - coin] + 1

B. Initialize minCoins with zeros for all amounts; update minCoins[i] by adding 1 for each coin

C. Use recursion without memoization to find minimum coins

D. Initialize minCoins with int.MaxValue for all amounts including 0; update minCoins[i] by subtracting coin values

Solution

1. Step 1: Proper initialization

   Set minCoins[0] = 0 because zero coins are needed for amount 0; others set to int.MaxValue to represent unreachable amounts initially.

2. Step 2: Update minCoins using dynamic programming

   For each amount i, check each coin less than or equal to i. If minCoins[i - coin] is reachable, update minCoins[i] to the minimum of current value and minCoins[i - coin] + 1.

3. Final Answer:

   Initialize minCoins with 0 for amount 0 and int.MaxValue for others; update minCoins[i] by checking coins <= i and minCoins[i - coin] + 1 -> Option A

4. Quick Check:

   DP with correct initialization and update finds minimum coins [OK]

Hint: Initialize 0 for amount 0, update with min of previous +1 [OK]

Common Mistakes:

• Initializing all minCoins to zero
• Using recursion without memoization causing inefficiency
• Incorrectly updating minCoins by subtraction