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;

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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]
        }
    }
}

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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];
}

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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];

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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
}

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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.