In this article, we will learn to resolve the Coin Change problem in Java by using a dynamic programming algorithm

## Problem

• Given a set of infinite coins

• Find the number of ways to making change for a specific amount of money, without considering the order of the coins

## Example

• Input: given a set of infinite coins {2, 3, 1}. Find the number of ways to making change for 4

• Expected output: 4

• Explanation: those 4 ways are {1, 1, 1, 1}, {1, 1, 2}, {1, 3} and {2, 2}

## Dynamic Programming Approach

• To making change for a value `j`, need to use coins with a value less than or equal to `j`

• For each coin in the set `c[i]`, calculate the ways of making change `w[j] = w[j] + w[j - c[i]]` with `c[i] <= j <= target`, `w = 1` is base case, `w[target]` is the final result

``````import java.util.Arrays;

public class DP_CoinChange {
static int countWays(int[] coins, int targetCoinChange) {
int[] wayOfCoinChanges = new int[targetCoinChange+1];

wayOfCoinChanges = 1;

for (int i = 0; i < coins.length; i++) {
for (int j = coins[i]; j <= targetCoinChange; j++) {
wayOfCoinChanges[j] += wayOfCoinChanges[j - coins[i]];
}
System.out.println(Arrays.toString(wayOfCoinChanges));
}

return wayOfCoinChanges[targetCoinChange];
}

public static void main(String[] args) {
System.out.println(countWays(new int[]{2, 3, 1}, 4));
}
}
``````
• Time complexity is O(nm) and space complexity is O(n) where n is the change target, m is the number of coins