Coin Change Total Number of Ways in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the ways array with 0.

DSA C

int ways[amount + 1] = {0};
ways[0] = [1];

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A1

B0

Camount

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting ways[0] to 0 will cause the algorithm to fail to count any combinations.

Using negative or amount value instead of 1.

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins.

2fill in blank

medium

Complete the code to update the ways array inside the nested loop.

DSA C

for (int i = 0; i < n; i++) {
    for (int j = coins[i]; j <= amount; j++) {
        ways[j] += [1];
    }
}

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Aways[j - coins[i]]

Bcoins[i]

Cj

Dways[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Adding coins[i] directly instead of ways[j - coins[i]].

Using ways[j] on the right side causes incorrect counting.

ways[j] is updated by adding ways[j - coins[i]] which represents the number of ways to make amount j - coins[i].

3fill in blank

hard

Fix the error in the loop condition to avoid out-of-bounds access.

DSA C

for (int j = coins[i]; j [1] amount; j++) {
    ways[j] += ways[j - coins[i]];
}

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A>

B<

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < causes the last amount to be skipped.

Using > or >= causes the loop to not run or run incorrectly.

The loop must run while j is less than or equal to amount to cover all amounts up to the target.

4fill in blank

hard

Fill both blanks to complete the function signature and return statement.

DSA C

int [1](int coins[], int n, int amount) {
    int ways[amount + 1] = {0};
    ways[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = coins[i]; j <= amount; j++) {
            ways[j] += ways[j - coins[i]];
        }
    }
    return [2];
}

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AcountWays

Bways[amount]

Camount

Dcoins

Attempts:

3 left

💡 Hint

Common Mistakes

Returning amount or coins instead of ways[amount].

Using a wrong function name that does not match the problem.

The function is named countWays and returns ways[amount], the total number of ways to make the amount.

5fill in blank

hard

Fill all three blanks to complete the main function that calls countWays and prints the result.

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount);

int main() {
    int coins[] = {1, 2, 5};
    int n = sizeof(coins) / sizeof(coins[0]);
    int amount = 5;
    int result = [1](coins, n, amount);
    printf("Total ways to make %d: %d\n", [2], [3]);
    return 0;
}

Drag options to blanks, or click blank then click option'

AcountWays

Bamount

Cresult

Dcoins

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names in printf.

Calling a wrong function or passing wrong arguments.

Call countWays with coins, n, amount; print amount and result to show total ways.

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