Coin Change Minimum Coins in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the minimum coins array with a large number.

DSA C

int minCoins[amount + 1];
for (int i = 0; i <= amount; i++) {
    minCoins[i] = [1];
}
minCoins[0] = 0;

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

Aamount + 1

B0

C-1

DINT_MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 will make all values zero and break the logic.

Using -1 is invalid for minimum calculation.

We initialize minCoins with amount + 1 as a large number to represent infinity since the maximum coins needed can't exceed amount.

2fill in blank

medium

Complete the code to loop through each coin for the current amount.

DSA C

for (int i = 1; i <= amount; i++) {
    for (int j = 0; j < [1]; j++) {
        if (coins[j] <= i) {
            // update minCoins[i]
        }
    }
}

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

Aamount

Bi

CnumCoins

Dcoins

Attempts:

3 left

💡 Hint

Common Mistakes

Using amount instead of numCoins causes out-of-bound errors.

Using coins as loop limit is invalid since coins is an array.

We loop through all coins using numCoins which is the count of coins available.

3fill in blank

hard

Fix the error in updating minCoins[i] to find the minimum coins needed.

DSA C

if (coins[j] <= i) {
    minCoins[i] = (minCoins[i] > minCoins[i - coins[j]] + 1) ? [1] : minCoins[i];
}

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

AminCoins[i - coins[j]] + 1

BminCoins[i] + 1

CminCoins[i - coins[j]]

Dcoins[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1 to minCoins[i] instead of minCoins[i - coins[j]] causes wrong results.

Using coins[j] directly is incorrect.

We update minCoins[i] with minCoins[i - coins[j]] + 1 because we add one coin to the minimum coins needed for the smaller amount.

4fill in blank

hard

Fill both blanks to return the correct result or -1 if no solution exists.

DSA C

return (minCoins[amount] == [1]) ? [2] : minCoins[amount];

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

Aamount + 1

B-1

C0

DINT_MAX

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 when no solution exists.

Comparing with INT_MAX instead of amount + 1.

If minCoins[amount] is still amount + 1, it means no solution was found, so return -1.

5fill in blank

hard

Fill all three blanks to complete the function signature and variable declarations.

DSA C

int [1](int* coins, int [2], int [3]) {
    int minCoins[amount + 1];
    // function body
}

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

AcoinChange

BnumCoins

Camount

DcoinsCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent parameter names.

Using coinsCount instead of numCoins.

The function is named coinChange, takes coins array, number of coins as numCoins, and amount as amount.

Practice

1. What does the Coin Change Minimum Coins problem ask you to find?

easy

A. The smallest number of coins needed to make a given amount

B. The total number of different coins available

C. The largest coin value in the set

D. The total sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of coins needed to make a specific amount using given coin denominations.

2. Step 2: Identify the correct description

   The smallest number of coins needed to make a given amount correctly describes this goal, while others describe unrelated concepts.

3. Final Answer:

   The smallest number of coins needed to make a given amount -> Option A

4. Quick Check:

   Minimum coins = smallest count needed [OK]

Hint: Focus on 'minimum coins' needed to form the amount [OK]

Common Mistakes:

• Confusing total coin count with minimum coins
• Thinking about coin values instead of counts
• Mixing up largest coin with minimum coins

2. Which of the following is the correct way to initialize the DP array for the Coin Change Minimum Coins problem in C?

easy

A. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = 0;

B. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = -1;

C. int dp[amount+1] = {0};

D. int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX;

Solution

1. Step 1: Understand DP initialization

   We initialize dp array with a large number (like INT_MAX) to represent impossible states initially.

2. Step 2: Identify correct initialization

   int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX; correctly sets all dp values to INT_MAX, except dp[0] which is usually set to 0 later.

3. Final Answer:

   int dp[amount+1]; for(int i=0; i<=amount; i++) dp[i] = INT_MAX; -> Option D

4. Quick Check:

   Initialize dp with max values for minimum search [OK]

Hint: Use INT_MAX to mark unreachable amounts initially [OK]

Common Mistakes:

• Initializing dp with zeros causes wrong minimum calculations
• Using -1 without handling leads to errors
• Not setting dp[0] = 0 explicitly

3. What is the output of the following code snippet for coins = {1, 3, 4} and amount = 6?

int coins[] = {1, 3, 4};
int amount = 6;
int dp[7];
for(int i=0; i<=amount; i++) dp[i] = amount+1;
dp[0] = 0;
for(int i=1; i<=amount; i++) {
  for(int j=0; j<3; j++) {
    if(coins[j] <= i) {
      dp[i] = dp[i] < dp[i - coins[j]] + 1 ? dp[i] : dp[i - coins[j]] + 1;
    }
  }
}
printf("%d\n", dp[amount]);

medium

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace dp array updates for amount 6

   Calculate minimum coins for each amount: - dp[0]=0 - dp[1]=1 (1 coin) - dp[2]=2 (1+1) - dp[3]=1 (3 coin) - dp[4]=1 (4 coin) - dp[5]=2 (4+1) - dp[6]=2 (3+3)

2. Step 2: Final dp[6] value

   dp[6] = 2, minimum coins needed to make 6.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Minimum coins for 6 = 2 [OK]

Hint: Trace dp array from 0 to amount stepwise [OK]

Common Mistakes:

• Forgetting to update dp[i] with minimum value
• Incorrectly handling coin indices
• Assuming dp[i] equals i always

4. Identify the bug in this code snippet for Coin Change Minimum Coins:

int coins[] = {1, 2, 5};
int amount = 11;
int dp[12];
for(int i=0; i<=amount; i++) dp[i] = 0;
dp[0] = 0;
for(int i=1; i<=amount; i++) {
  for(int j=0; j<3; j++) {
    if(coins[j] <= i) {
      dp[i] = dp[i - coins[j]] + 1;
    }
  }
}
printf("%d\n", dp[amount]);

medium

A. dp array is not initialized with large values

B. dp[0] should be set to 1

C. The inner loop should run from j=1 to 3

D. The condition coins[j] <= i is incorrect

Solution

1. Step 1: Check dp initialization

   dp array is initialized with zeros, which incorrectly suggests zero coins needed for all amounts.

2. Step 2: Correct initialization needed

   dp should be initialized with a large number (like INT_MAX) to represent unreachable states, then updated with minimum coins.

3. Final Answer:

   dp array is not initialized with large values -> Option A

4. Quick Check:

   Initialize dp with max values to find minimum [OK]

Hint: Initialize dp with large values, not zeros [OK]

Common Mistakes:

• Setting dp to zero causes wrong minimum calculations
• Not updating dp[i] with minimum of previous values
• Incorrect loop ranges

5. Given coins = {2, 5, 10} and amount = 7, what is the minimum number of coins needed? If it is not possible, return -1. Use the Coin Change Minimum Coins logic.

hard

A. -1

B. 2

C. 3

D. 1

Solution

1. Step 1: Check possible combinations for amount 7

   Coins are 2, 5, 10. Possible sums: - 2+5=7 (2 coins) - 5+2=7 (2 coins) - 10 is too large So minimum coins = 2.

2. Step 2: Confirm no smaller combination

   Since 1-coin denominations are not available, 2 coins is minimum.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   7 = 2 + 5, minimum coins = 2 [OK]

Hint: Try all coin combos to find minimum or detect impossibility [OK]

Common Mistakes:

• Assuming 7 cannot be formed without 1-coin denomination
• Returning -1 incorrectly
• Ignoring order of coins