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 <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1];
    for (int i = 0; i <= amount; i++) dp[i] = 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 n = 3;
    int amount = 5;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A 7

B 6

C 5

D 8

💡 Hint

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

The code uses dynamic programming to count the number of ways to make amount 5 using coins 1, 2, and 3. The total ways are 7.

❓ Predict Output

intermediate

Output of Coin Change Ways with Different Coins

What is the output of the following C code that calculates the total number of ways to make change for amount 4 using coins {1, 3, 4}?

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount+1];
    for (int i = 0; i <= amount; i++) dp[i] = 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, 3, 4};
    int n = 3;
    int amount = 4;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A 4

B 5

C 6

D 3

💡 Hint

Count all combinations of coins 1, 3, and 4 that sum to 4.

The code counts ways to make 4 using coins 1, 3, and 4. The total ways are 3.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does the following C code produce when trying to calculate the number of ways to make change for amount 5 using coins {1, 2}?

DSA C

#include <stdio.h>

int countWays(int coins[], int n, int amount) {
    int dp[amount];
    for (int i = 0; i <= amount; i++) dp[i] = 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};
    int n = 2;
    int amount = 5;
    printf("%d\n", countWays(coins, n, amount));
    return 0;
}

A Compilation error: array size must be constant

B Segmentation fault (out of bounds access)

C Logical error: returns 0 instead of correct count

D No error, outputs correct result

💡 Hint

Check the size of the dp array and the indices used.

The dp array is declared with size amount (5), but code accesses dp[amount] (dp[5]) which is out of bounds causing segmentation fault.

🧠 Conceptual

advanced

Understanding DP Array Initialization in Coin Change

Why is dp[0] initialized to 1 in the dynamic programming solution for the coin change total number of ways problem?

A Because dp[0] is initialized to 1 to avoid division by zero

B Because dp[0] stores the total number of coins used

C Because there is exactly one way to make amount 0: by choosing no coins

D Because dp[0] is used to store the maximum coin value

💡 Hint

Think about how many ways there are to make zero amount.

dp[0] = 1 means there is one way to make amount zero: by not choosing any coin.

🚀 Application

expert

Number of Ways to Make Change for Large Amount

Given coins {1, 5, 10, 25} and amount 100, what is the number of ways to make change using the standard dynamic programming approach?

A 242

B 292

C 255

D 300

💡 Hint

This is a classic coin change problem; the answer is a known value.

The number of ways to make change for 100 cents using coins {1,5,10,25} is 242.