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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with zeros.

DSA C++

int dp[[1] + 1] = {0};

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

Aamount

Bcoins.size()

Ccoins.size() + 1

Damount + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using coins.size() instead of amount for dp array size.

Forgetting to add 1 to amount for array size.

We need an array of size amount + 1 to store ways to make each amount from 0 to amount.

2fill in blank

medium

Complete the code to set the base case for dp array.

DSA C++

dp[[1]] = 1;

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

A1

Bcoins[0]

C0

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[1] = 1 instead of dp[0].

Setting dp[amount] = 1 incorrectly.

There is exactly one way to make amount 0: by choosing no coins.

3fill in blank

hard

Fix the error in the loop condition to iterate over amounts correctly.

DSA C++

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

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

A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using < causes last amount to be skipped.

Using > or >= causes incorrect loop behavior.

We need to include amount itself, so the loop should run while j <= amount.

4fill in blank

hard

Fill both blanks to update dp array with ways including current coin.

DSA C++

dp[j] [1]= dp[j - coins[i]];

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

A/

B+

C*

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using -= or *= instead of += causes wrong results.

Using / operator causes runtime error.

We add ways to make amount j - coins[i] to dp[j] to include current coin.

5fill in blank

hard

Fill all three blanks to complete the coin change function.

DSA C++

int change(int amount, vector<int>& coins) {
    vector<int> dp(amount + 1, 0);
    dp[[1]] = 1;
    for (int i = 0; i [2] coins.size(); i++) {
        for (int j = coins[i]; j [3] amount; j++) {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[amount];
}

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

A0

B<

C<=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Wrong base case index.

Using <= in outer loop causing out-of-bounds.

Using < in inner loop causing last amount to be skipped.

dp[0] = 1 as base case; outer loop runs while i < coins.size(); inner loop runs while j <= amount.

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