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 minimum number of coins 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 minimum number of coins of making change for 3

Expected output: 1

Explanation: there're 3 ways to making change for 3: {3}, {1, 1, 1}, {1, 2}, minimum is {3}

## Dynamic Programming Approach

To making change for a value

`i`

, need to use coins with a value less than or equal to`i`

For each coin in the set

`c[j]`

, calculate the minimum number of coins to making change`m[i] = Math.min(m[i], m[i - c[j]]) + 1`

with`c[j] <= i <= target`

,`m[0] = 0`

is base case,`m[target]`

is the final result

```
import java.util.Arrays;
public class DP_CoinChange2 {
static int countMin(int[] coins, int targetCoinChange) {
int[] minNoOfCoins = new int[targetCoinChange+1];
Arrays.fill(minNoOfCoins, targetCoinChange + 1);
minNoOfCoins[0] = 0;
for (int i = 1; i <= targetCoinChange; i++) {
for (int j = 0; j < coins.length; j++) {
if (coins[j] <= i) {
minNoOfCoins[i] = Math.min(minNoOfCoins[i], minNoOfCoins[i - coins[j]] + 1);
}
}
}
return minNoOfCoins[targetCoinChange] > targetCoinChange ? -1 : minNoOfCoins[targetCoinChange];
}
public static void main(String[] args) {
System.out.println(countMin(new int[]{2, 3, 1}, 3));
}
}
```

- Time complexity is O(nm) and space complexity is O(n) where n is the change target, m is the number of coins