Coin Change Minimum Coins in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of minimum coins for amount 11

What is the output of the following C# code that calculates the minimum number of coins needed to make amount 11 using coins {1, 2, 5}?

DSA C#

using System;
class Program {
    static int MinCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        for (int i = 1; i <= amount; i++) dp[i] = max;
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            foreach (int coin in coins) {
                if (coin <= i) {
                    dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
    static void Main() {
        int[] coins = {1, 2, 5};
        int amount = 11;
        Console.WriteLine(MinCoins(coins, amount));
    }
}

A 3

B 4

C 5

D -1

💡 Hint

Think about the fewest coins to make 11 using 5, 2, and 1.

The minimum coins to make 11 are 5 + 5 + 1 = 3 coins.

❓ Predict Output

intermediate

Minimum coins for amount 7 with coins {2, 4}

What is the output of this C# code that finds the minimum coins needed to make amount 7 using coins {2, 4}?

DSA C#

using System;
class Program {
    static int MinCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount + 1];
        for (int i = 1; i <= amount; i++) dp[i] = max;
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            foreach (int coin in coins) {
                if (coin <= i) {
                    dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
    static void Main() {
        int[] coins = {2, 4};
        int amount = 7;
        Console.WriteLine(MinCoins(coins, amount));
    }
}

A 3

B 2

C -1

D 4

💡 Hint

Check if 7 can be formed by sums of 2 and 4.

7 cannot be formed by any combination of 2 and 4 coins, so output is -1.

🔧 Debug

advanced

Identify the error in coin change code

What error will this C# code produce when trying to find minimum coins for amount 5 with coins {1, 3, 4}?

DSA C#

using System;
class Program {
    static int MinCoins(int[] coins, int amount) {
        int max = amount + 1;
        int[] dp = new int[amount];
        for (int i = 1; i <= amount; i++) dp[i] = max;
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            foreach (int coin in coins) {
                if (coin <= i) {
                    dp[i] = Math.Min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
    static void Main() {
        int[] coins = {1, 3, 4};
        int amount = 5;
        Console.WriteLine(MinCoins(coins, amount));
    }
}

A ArgumentNullException

B IndexOutOfRangeException

C No error, output is 2

D InvalidOperationException

💡 Hint

Check array size and indexing in dp array.

dp array size is amount (5), but code accesses dp[5], which is out of range (max index 4).

🧠 Conceptual

advanced

Why use dynamic programming for coin change?

Why is dynamic programming the best approach to solve the minimum coin change problem?

A Because it stores results of subproblems to avoid repeated calculations

B Because it sorts coins and picks the smallest coin repeatedly

C Because it uses a greedy approach that always picks the largest coin first

D Because it tries all combinations recursively without storing results

💡 Hint

Think about avoiding repeated work in overlapping subproblems.

Dynamic programming stores results of smaller amounts to build up solutions efficiently.

🚀 Application

expert

Minimum coins for amount 27 with coins {1, 5, 10, 25}

What is the minimum number of coins needed to make amount 27 using coins {1, 5, 10, 25}?

A 4

B 6

C 5

D 3

💡 Hint

Try to use the largest coins first and then smaller coins.

The minimum is 3 coins: 25 + 1 + 1.

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