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

Coin Change Total Number of Ways

📖 Scenario: You have a collection of coins of different values. You want to find out how many different ways you can make a certain amount of money using these coins. This is like figuring out how many different combinations of coins can add up to the total amount.

🎯 Goal: Build a program that calculates the total number of ways to make a target amount using given coin values.

📋 What You'll Learn

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

Create an integer variable called amount and set it to 5

Create an integer array called ways to store the number of ways to make each amount from 0 to amount

Use a loop to update the ways array based on each coin

Print the total number of ways to make the amount

💡 Why This Matters

🌍 Real World

This problem is useful in finance and vending machines where you need to know how many ways to give change.

💼 Career

Understanding this algorithm helps in roles involving dynamic programming, problem solving, and optimization.

1

Create the coins array and amount variable

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

DSA C

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

Hint

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

2

Create the ways array and initialize it

Create an integer array called ways with size amount + 1. Initialize all elements to 0 except ways[0] which should be set to 1.

DSA C

#include 

int main() {
    int coins[] = {1, 2, 5};
    int amount = 5;

    // Create and initialize the ways array
    // Your code here

    return 0;
}

Hint

Use int ways[amount + 1] = {0}; to create the array and then set ways[0] = 1;.

3

Update ways array using coins

Use a for loop with variable i from 0 to 2 to iterate over coins. Inside it, use another for loop with variable j from coins[i] to amount. Update ways[j] by adding ways[j - coins[i]].

DSA C

#include 

int main() {
    int coins[] = {1, 2, 5};
    int amount = 5;
    int ways[6] = {0};
    ways[0] = 1;

    // Update ways array using coins
    // Your code here

    return 0;
}

Hint

Use nested loops: outer loop over coins with i, inner loop over amounts with j. Update ways[j] by adding ways[j - coins[i]].

4

Print the total number of ways

Print the value of ways[amount] using printf with the format "%d\n".

DSA C

#include 

int main() {
    int coins[] = {1, 2, 5};
    int amount = 5;
    int ways[6] = {0};
    ways[0] = 1;

    for (int i = 0; i < 3; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            ways[j] += ways[j - coins[i]];
        }
    }

    // Print the total number of ways
    // Your code here

    return 0;
}

Hint

Use printf("%d\n", 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 number of ways to make a target amount using given 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 sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible ways to combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, not minimum coins.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

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 it finds maximum coin value
• Assuming it sums coin denominations

2. Which of the following is the correct way to declare the DP array for the Coin Change Total Number of Ways problem in C?

easy

A. int dp[target];

B. int dp[target+1];

C. int dp[target+1][coins];

D. int dp[coins+1];

Solution

1. Step 1: Understand DP array size

   We need to store ways for amounts from 0 to target, so size is target+1.

2. Step 2: Choose correct declaration

   One-dimensional array of size target+1 is used to store ways for each amount.

3. Final Answer:

   int dp[target+1]; -> Option B

4. Quick Check:

   Array size = target+1 [OK]

Hint: DP array size must cover 0 to target inclusive [OK]

Common Mistakes:

• Using dp[target] misses amount 0
• Using 2D array unnecessarily
• Using dp[coins+1] which is unrelated

3. What is the output of the following C code snippet for coins = {1, 2} and target = 3?

int coins[] = {1, 2};
int target = 3;
int dp[4] = {1, 0, 0, 0};
for (int i = 0; i < 2; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. 1

B. 3

C. 2

D. 4

Solution

1. Step 1: Initialize dp array

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

2. Step 2: Process coin 1

   For j=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 j=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 C

5. Quick Check:

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

Hint: Add ways from smaller amounts using each coin [OK]

Common Mistakes:

• Not initializing dp[0] to 1
• Updating dp array incorrectly
• Mixing order of loops

4. Identify the error in this code snippet for calculating total ways to make change:

int coins[] = {1, 2, 3};
int target = 4;
int dp[5] = {0};
dp[0] = 1;
for (int i = 0; i <= 3; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. Loop i should run from 0 to 2, not 0 to 3.

B. dp array size is too small.

C. dp[0] should be initialized to 0.

D. Inner loop should start from 0, not coins[i].

Solution

1. Step 1: Check loop bounds for coins array

   coins has 3 elements indexed 0 to 2, but loop runs i=0 to 3 (4 iterations) causing out-of-bounds access.

2. Step 2: Correct loop range

   Change loop to i=0 to 2 to avoid accessing coins[3] which is invalid.

3. Final Answer:

   Loop i should run from 0 to 2, not 0 to 3. -> Option A

4. Quick Check:

   Array index out-of-bounds error [OK]

Hint: Loop over coins must match array length exactly [OK]

Common Mistakes:

• Using wrong loop bounds causing crashes
• Incorrect dp initialization
• Starting inner loop incorrectly

5. Given coins = {1, 3, 5} and target = 7, how many total ways are there to make change? Use the bottom-up DP approach.

hard

A. 6

B. 5

C. 7

D. 4

Solution

1. Step 1: Initialize dp array

   dp[0]=1, dp[1..7]=0 initially.

2. Step 2: Process coin 1

   Update dp for amounts 1 to 7 by adding dp[j-1]: dp=[1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   Update dp for j=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

   Update dp for j=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 D

6. Quick Check:

   Count ways with coins {1,3,5} for 7 = 4 [OK]

Hint: Add ways for each coin incrementally [OK]

Common Mistakes:

• Not updating dp correctly for each coin
• Forgetting dp[0] = 1 initialization
• Mixing order of coin processing