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 <iostream>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1] = {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 amount = 5;
    int ways = countWays(coins, 3, amount);
    std::cout << ways << std::endl;
    return 0;
}

A 5

B 7

C 6

D 8

💡 Hint

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

The code uses dynamic programming to count the number of combinations (order doesn't matter) to make change. For amount 5 with coins {1, 2, 3}, the total ways are 5.

❓ Predict Output

intermediate

Output of Coin Change with Single Coin

What is the output of this code that calculates the number of ways to make change for amount 4 using only coin {2}?

DSA C++

#include <iostream>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1] = {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[] = {2};
    int amount = 4;
    int ways = countWays(coins, 1, amount);
    std::cout << ways << std::endl;
    return 0;
}

A 1

B 2

C 0

D 3

💡 Hint

How many ways can you make 4 using only 2s?

Using only coin 2, amount 4 can be made in exactly one combination: {2,2}.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does the following code produce when trying to calculate the number of ways to make change for amount 3 with coins {1, 2}?

DSA C++

#include <iostream>

int countWays(int coins[], int n, int amount) {
    int dp[amount];
    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 amount = 3;
    int ways = countWays(coins, 2, amount);
    std::cout << ways << std::endl;
    return 0;
}

A No error, outputs 3

B Compilation error: dp array not initialized

C Logical error: returns 0 instead of correct count

D Segmentation fault due to out-of-bounds access

💡 Hint

Check the size of dp array and indexing.

The dp array is declared with size amount (3), but code accesses dp[3], which is out of bounds causing segmentation fault.

❓ Predict Output

advanced

Output of Coin Change with Zero Amount

What is the output of this code that calculates the number of ways to make change for amount 0 with coins {1, 2, 3}?

DSA C++

#include <iostream>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1] = {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 amount = 0;
    int ways = countWays(coins, 3, amount);
    std::cout << ways << std::endl;
    return 0;
}

A 0

B 1

C Undefined behavior

D 3

💡 Hint

How many ways to make amount zero?

There is exactly one way to make amount zero: use no coins.

❓ Predict Output

expert

Output of Coin Change with Large Amount and Multiple Coins

What is the output of this code that calculates the number of ways to make change for amount 10 using coins {2, 5, 3, 6}?

DSA C++

#include <iostream>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1] = {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[] = {2, 5, 3, 6};
    int amount = 10;
    int ways = countWays(coins, 4, amount);
    std::cout << ways << std::endl;
    return 0;
}

A 7

B 4

C 5

D 6

💡 Hint

Count all combinations of coins {2,3,5,6} that sum to 10.

The code counts all combinations without order. For amount 10 and coins {2,5,3,6}, total ways are 5.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?
Given coins and a total amount, how many ways can you make that amount?

easy

A. The total number of ways to make the amount using given coins

B. The minimum number of coins needed to make the amount

C. The maximum coin value used to make the amount

D. The total sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the count of different ways to combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   This is not about minimum coins or maximum coin value, but total ways to form the amount.

3. Final Answer:

   The total number of ways to make the amount using given coins -> Option A

4. Quick Check:

   Count ways, not minimum or max coin [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins needed
• Thinking it's about maximum coin value
• Mixing sum of coins with number of ways

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

int amount = 5;
vector<int> dp(amount + 1, 0);
dp[0] = ?;

easy

A. 1

B. 0

C. -1

D. amount

Solution

1. Step 1: Understand base case initialization

   dp[0] represents ways to make amount 0, which is 1 (using no coins).

2. Step 2: Set dp[0] to 1

   Setting dp[0] = 1 allows building solutions for other amounts correctly.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   dp[0] = 1 means one way to make zero amount [OK]

Hint: dp[0] always 1 for coin change ways [OK]

Common Mistakes:

• Setting dp[0] to 0 causes no valid ways
• Using negative values for dp initialization
• Confusing dp array meaning with minimum coins

3. What is the output of the following C++ code?

#include <iostream>
#include <vector>
using namespace std;

int main() {
    vector<int> coins = {1, 2};
    int amount = 3;
    vector<int> dp(amount + 1, 0);
    dp[0] = 1;
    for (int coin : coins) {
        for (int x = coin; x <= amount; x++) {
            dp[x] += dp[x - coin];
        }
    }
    cout << dp[amount] << endl;
    return 0;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Trace dp array updates for coins {1, 2} and amount 3

   Initially dp = [1,0,0,0]. After coin=1: dp = [1,1,1,1]. After coin=2: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2, so dp = [1,1,2,2].

2. Step 2: Final dp[3] value

   dp[3] = 2 means there are 2 ways to make amount 3 with coins 1 and 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

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

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

Common Mistakes:

• Mixing order of loops causing wrong counts
• Not initializing dp[0] to 1
• Confusing dp[x] update with subtraction

4. Find the error in this C++ code snippet for Coin Change Total Number of Ways:

vector<int> dp(amount + 1, 0);
dp[0] = 1;
for (int x = 1; x <= amount; x++) {
    for (int coin : coins) {
        if (coin <= x) {
            dp[x] += dp[x - coin];
        }
    }
}

medium

A. The dp array size is too small

B. dp[0] should be initialized to 0

C. The condition if (coin <= x) is incorrect

D. The outer loop should iterate over coins, inner over amounts

Solution

1. Step 1: Analyze loop order for correct counting

   Outer loop must be over coins to avoid counting duplicates; current code counts duplicates.

2. Step 2: Correct loop order

   Change outer loop to for (int coin : coins) and inner loop to for (int x = coin; x <= amount; x++) to count unique combinations.

3. Final Answer:

   The outer loop should iterate over coins, inner over amounts -> Option D

4. Quick Check:

   Loop order affects counting unique ways [OK]

Hint: Outer loop coins, inner loop amounts for unique ways [OK]

Common Mistakes:

• Swapping loops causes counting duplicates
• Incorrect dp[0] initialization
• Wrong dp array size causing out-of-bounds

5. You have coins {1, 3, 5} and want to find the total number of ways to make amount 7. Which approach correctly computes this using dynamic programming?

hard

A. Sort coins descending and greedily subtract largest coin until amount is zero

B. Use a 2D dp array with rows as coins and columns as amounts, fill with minimum coins needed

C. Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount

D. Use recursion without memoization to count ways

Solution

1. Step 1: Identify correct DP approach for total ways

   Using 1D dp array with dp[0]=1 and updating dp[x] for each coin counts total ways efficiently.

2. Step 2: Reject other approaches

   2D dp for minimum coins is different problem; greedy fails for total ways; recursion without memo is inefficient.

3. Final Answer:

   Use a 1D dp array, initialize dp[0]=1, then for each coin, update dp[x] += dp[x - coin] for x from coin to amount -> Option C

4. Quick Check:

   1D dp with coin loops counts ways correctly [OK]

Hint: 1D dp with coin loops counts ways, not min coins [OK]

Common Mistakes:

• Confusing total ways with minimum coins problem
• Using greedy which fails for counting ways
• Ignoring memoization causing slow recursion