Coin Change Total Number of Ways in DSA C# - Build from Scratch

Coin Change Total Number of Ways

📖 Scenario: You are helping a cashier machine figure out how many different ways it can give change for a certain amount using given coin denominations.

🎯 Goal: Build a program that calculates the total number of ways to make change for a given amount using a list of coin denominations.

📋 What You'll Learn

Create an integer array called coins with the exact values 1, 2, 5

Create an integer variable called amount and set it to 5

Create an integer array called ways with size amount + 1 to store ways to make each amount

Initialize ways[0] to 1 because there is one way to make amount zero

Use a for loop with variable coin iterating over coins

Inside that, use a for loop with variable i from coin to amount

Update ways[i] by adding ways[i - coin]

Print the total number of ways to make change for amount using ways[amount]

💡 Why This Matters

🌍 Real World

Cashiers and vending machines use similar logic to calculate how to give change in different ways.

💼 Career

Understanding this problem helps in roles involving algorithms, software development, and financial software where optimization and counting combinations are important.

1

Create the coin denominations array and amount

Create an integer array called coins with the exact values 1, 2, 5 and an integer variable called amount set to 5.

DSA C#

# Create the coins array and amount variable
# Your code here

Hint

Use int[] coins = {1, 2, 5}; to create the array and int amount = 5; to set the amount.

2

Create and initialize the ways array

Create an integer array called ways with size amount + 1 and set ways[0] to 1.

DSA C#

int[] coins = {1, 2, 5};
int amount = 5;
# Create the ways array and initialize ways[0] to 1
# Your code here

Hint

Use new int[amount + 1] to create the array and set ways[0] = 1; because there is one way to make zero amount.

3

Calculate the total number of ways using nested loops

Use a for loop with variable coin iterating over coins. Inside it, use a for loop with variable i from coin to amount. Update ways[i] by adding ways[i - coin].

DSA C#

int[] coins = {1, 2, 5};
int amount = 5;
int[] ways = new int[amount + 1];
ways[0] = 1;
# Use nested loops to update ways array
# Your code here

Hint

Use foreach (int coin in coins) and inside it a for loop from coin to amount. Update ways[i] by adding ways[i - coin].

4

Print the total number of ways to make change

Print the total number of ways to make change for amount using ways[amount].

DSA C#

int[] coins = {1, 2, 5};
int amount = 5;
int[] ways = new int[amount + 1];
ways[0] = 1;

foreach (int coin in coins)
{
    for (int i = coin; i <= amount; i++)
    {
        ways[i] += ways[i - coin];
    }
}
# Print the total ways to make change
# Your code here

Hint

Use System.Console.WriteLine(ways[amount]); to print the result.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total sum of all coin denominations.

B. The minimum number of coins needed to make a target amount.

C. The maximum value of coins that can be used to reach the target.

D. The total number of ways to make a target amount using given coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible combinations of coins that sum up to the target amount.

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 target amount using given coin denominations. -> Option D

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking about maximum coin value instead of count
• Ignoring that order of coins does not matter

2. Which of the following C# 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[5]; dp[1] = 1;

D. int[] dp = new int[6]; 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 there is 1 way to make amount 0 (using no coins).

3. Final Answer:

   int[] dp = new int[6]; dp[0] = 1; -> Option A

4. Quick Check:

   Array size = amount+1, dp[0] = 1 [OK]

Hint: DP size = amount + 1; dp[0] = 1 base case [OK]

Common Mistakes:

• Using 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 C# code snippet?

int[] coins = {1, 2};
int amount = 3;
int[] dp = new int[amount + 1];
dp[0] = 1;
foreach (int coin in coins) {
  for (int i = coin; i <= amount; i++) {
    dp[i] += dp[i - coin];
  }
}
Console.WriteLine(dp[amount]);

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] initially, dp[0]=1 means 1 way to make 0.

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. Final Answer:

   2 -> Option A

5. Quick Check:

   Ways to make 3 with coins 1,2 = 2 [OK]

Hint: Add dp[i-coin] for each coin and amount [OK]

Common Mistakes:

• Mixing up dp array updates order
• Forgetting to initialize dp[0] = 1
• Miscounting ways by double counting

4. Identify the error in the following C# code for Coin Change Total Number of Ways and select the fix.

int[] coins = {1, 2, 3};
int amount = 4;
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int i = 0; i <= amount; i++) {
  for (int coin : coins) {
    if (i >= coin) {
      dp[i] += dp[i - coin];
    }
  }
}
Console.WriteLine(dp[amount]);

medium

A. Change dp[0] = 1 to dp[1] = 1

B. Change outer loop to 'for (int i = 1; i <= amount; i++)' and inner loop to 'for (int coin in coins)'

C. Replace 'for (int coin : coins)' with 'foreach (int coin in coins)'

D. No error, code is correct

Solution

1. Step 1: Identify incorrect loop syntax

   C# does not support 'for (int coin : coins)', it uses 'foreach (int coin in coins)'.

2. Step 2: Correct the loop syntax

   Replace 'for (int coin : coins)' with 'foreach (int coin in coins)' to fix syntax error.

3. Final Answer:

   Replace 'for (int coin : coins)' with 'foreach (int coin in coins)' -> Option C

4. Quick Check:

   C# uses foreach for collections [OK]

Hint: Use foreach for arrays in C# [OK]

Common Mistakes:

• Using Java-style for-each syntax in C#
• Changing dp base case incorrectly
• Assuming code is correct without checking syntax

5. Given coins = {1, 3, 5} and amount = 7, how many total ways are there to make amount 7? Use the coin change total number of ways logic.

hard

A. 5

B. 4

C. 7

D. 6

Solution

1. Step 1: Initialize dp array

   dp size = 8, dp[0] = 1, dp = [1,0,0,0,0,0,0,0]

2. Step 2: Process coin 1

   Update dp for i=1 to 7: dp[i]+=dp[i-1]; dp becomes [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:

   Ways to make 7 with coins 1,3,5 = 4 [OK]

Hint: Update dp for each coin from coin to amount [OK]

Common Mistakes:

• Miscounting ways by skipping coins
• Not updating dp array correctly
• Confusing total ways with minimum coins