Coin Change Total Number of Ways in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of the following C code that calculates the total number of ways to make change for amount 5 using coins {1, 2, 3}?

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1];
    for (int i = 0; i <= amount; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}

int main() {
    int coins[] = {1, 2, 3};
    int n = 3;
    int amount = 5;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A 7

B 6

C 5

D 8

💡 Hint

Think about how many combinations of coins 1, 2, and 3 can sum to 5.

The code uses dynamic programming to count the number of ways to make amount 5 using coins 1, 2, and 3. The total ways are 7.

❓ Predict Output

intermediate

Output of Coin Change Ways with Different Coins

What is the output of the following C code that calculates the total number of ways to make change for amount 4 using coins {1, 3, 4}?

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1];
    for (int i = 0; i <= amount; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}

int main() {
    int coins[] = {1, 3, 4};
    int n = 3;
    int amount = 4;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A 4

B 5

C 6

D 3

💡 Hint

Count all combinations of coins 1, 3, and 4 that sum to 4.

The code counts ways to make 4 using coins 1, 3, and 4. The total ways are 3.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does the following C code produce when trying to calculate the number of ways to make change for amount 5 using coins {1, 2}?

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount];
    for (int i = 0; i <= amount; i++) dp[i] = 0;
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}

int main() {
    int coins[] = {1, 2};
    int n = 2;
    int amount = 5;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A Compilation error: array size must be constant

B Segmentation fault (out of bounds access)

C Logical error: returns 0 instead of correct count

D No error, outputs correct result

💡 Hint

Check the size of the dp array and the indices used.

The dp array is declared with size amount (5), but code accesses dp[amount] (dp[5]) which is out of bounds causing segmentation fault.

🧠 Conceptual

advanced

Understanding DP Array Initialization in Coin Change

Why is dp[0] initialized to 1 in the dynamic programming solution for the coin change total number of ways problem?

A Because dp[0] is initialized to 1 to avoid division by zero

B Because dp[0] stores the total number of coins used

C Because there is exactly one way to make amount 0: by choosing no coins

D Because dp[0] is used to store the maximum coin value

💡 Hint

Think about how many ways there are to make zero amount.

dp[0] = 1 means there is one way to make amount zero: by not choosing any coin.

🚀 Application

expert

Number of Ways to Make Change for Large Amount

Given coins {1, 5, 10, 25} and amount 100, what is the number of ways to make change using the standard dynamic programming approach?

A 242

B 292

C 255

D 300

💡 Hint

This is a classic coin change problem; the answer is a known value.

The number of ways to make change for 100 cents using coins {1,5,10,25} is 242.

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