Ways To Make Coin Change

Given an infinite supply of coins of varying denominations, determine the total number of ways to make change for a specified value using these coins. If it's not possible to make the change, the result should be 0.

Input:

The first line contains an integer N, denoting the total number of denominations. The second line contains N space-separated integers, each representing a denomination value. The third line contains the integer V, representing the value to make change for.

Output:

An integer denoting the total number of ways to make change for V.

Example:

Input:

N = 3 D = [1, 2, 3] V = 4

Output:

4

Explanation:

There are four ways to make change for 4 using the denominations [1, 2, 3]:
1. 1+1+1+1
2. 1+1+2
3. 1+3
4. 2+2

Constraints:

1

Question

Ways To Make Coin Change

Given an infinite supply of coins of varying denominations, determine the total number of ways to make change for a specified value using these coins. If it's not possible to make the change, the result should be 0.

Input:

The first line contains an integer N, denoting the total number of denominations.
The second line contains N space-separated integers, each representing a denomination value.
The third line contains the integer V, representing the value to make change for.

Output:

An integer denoting the total number of ways to make change for V.

Example:

Input:

N = 3
D = [1, 2, 3]
V = 4

Output:

Explanation:

There are four ways to make change for 4 using the denominations [1, 2, 3]:
1. 1+1+1+1
2. 1+1+2
3. 1+3
4. 2+2

Constraints:

1 <= N <= 10
1 <= D[i] <= 10^5
1 <= V <= 2 * 10^3

Note: You don't need to handle the output; just implement the function as specified.

Accepted Answer

public class CoinChangeWays {

public static void main(String[] args) {

int[] denominations = {1, 2, 5};

int value = 5;

int ways = countWaysToMakeChange(denominations, value);

System.out.println("Total number of ways to make change: " + ways);

}

public static int countWaysToMakeChange(int[] denominations, int value) {

int[] dp = new int[value + 1];

dp[0] = 1; // There is one way to make change for 0

for (int coin : denominations) {

for (int i = coin; i <= value; i++) {

dp[i] += dp[i - coin];

}

return dp[value];

}

Accepted Answer

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