# HelloKoding

Practical coding guides

# Coin Change Problems

In this article, you will learn to analyze and resolve the Coin change problems by using a Dynamic programming algorithm

## Problem 1: Count ways of coin change

• 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 1

• 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}

## Approach 1: Dynamic programming

• 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

DP_CoinChange.java

``````package com.hellokoding.algorithm;

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: O(nm) with n is the change target, m is the number of coins
• Space complexity: O(n) with n is the change target

## Problem 2: Minimum coin change

• 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 2

• 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}

## Approach 2: Dynamic programming

• 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` is base case, `m[target]` is the final result

DP_CoinChange2.java

``````package com.hellokoding.algorithm;

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;

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: O(nm) with n is the change target, m is the number of coins
• Space complexity: O(n) with n is the change target