Coin Change Minimum Coins in DSA C# - Build from Scratch

Coin Change Minimum Coins

📖 Scenario: You are helping a cashier who needs to give change using the fewest coins possible. The cashier has coins of certain fixed values. Your task is to find the minimum number of coins needed to make a specific amount.

🎯 Goal: Build a program that calculates the minimum number of coins needed to make a given amount using a list of coin values.

📋 What You'll Learn

Create an array of coin values

Create a variable for the amount to make change for

Implement the logic to find the minimum number of coins needed

Print the minimum number of coins required

💡 Why This Matters

🌍 Real World

Cashiers and vending machines use similar logic to give change efficiently.

💼 Career

Understanding dynamic programming and optimization problems is useful for software engineers working on finance, retail, and algorithm design.

1

Create the coin values array

Create an integer array called coins with these exact values: 1, 3, and 5.

DSA C#

# Your code here

Hint

Use int[] coins = {1, 3, 5}; to create the array.

2

Set the amount to make change for

Create an integer variable called amount and set it to 11.

DSA C#

int[] coins = {1, 3, 5};
# Your code here

Hint

Use int amount = 11; to set the amount.

3

Calculate the minimum number of coins

Create an integer array called minCoins with size amount + 1. Initialize minCoins[0] to 0. Use a for loop with variable i from 1 to amount to fill minCoins[i] with the minimum coins needed. Inside this loop, use another for loop with variable j to iterate over coins. Update minCoins[i] if coins[j] is less than or equal to i and minCoins[i - coins[j]] + 1 is less than current minCoins[i].

DSA C#

int[] coins = {1, 3, 5};
int amount = 11;
# Your code here

Hint

Use dynamic programming with two nested loops to fill minCoins.

4

Print the minimum number of coins needed

Print the value of minCoins[amount] using Console.WriteLine.

DSA C#

int[] coins = {1, 3, 5};
int amount = 11;
int[] minCoins = new int[amount + 1];
minCoins[0] = 0;
for (int i = 1; i <= amount; i++)
{
    minCoins[i] = int.MaxValue;
    for (int j = 0; j < coins.Length; j++)
    {
        if (coins[j] <= i && minCoins[i - coins[j]] != int.MaxValue && minCoins[i - coins[j]] + 1 < minCoins[i])
        {
            minCoins[i] = minCoins[i - coins[j]] + 1;
        }
    }
}
# Your code here

Hint

The minimum number of coins to make 11 with coins 1, 3, and 5 is 3.

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