# 518. Coin Change II

## Description

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

Example 1:

Input: amount = 5, coins = [1,2,5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1


Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.


Example 3:

Input: amount = 10, coins = [10]
Output: 1


Constraints:

• 1 <= coins.length <= 300
• 1 <= coins[i] <= 5000
• All the values of coins are unique.
• 0 <= amount <= 5000

## Solutions

Dynamic programming.

Complete knapsack problem.

• class Solution {
public int change(int amount, int[] coins) {
int[] dp = new int[amount + 1];
dp[0] = 1;
for (int coin : coins) {
for (int j = coin; j <= amount; j++) {
dp[j] += dp[j - coin];
}
}
return dp[amount];
}
}


• class Solution {
public:
int change(int amount, vector<int>& coins) {
vector<int> dp(amount + 1);
dp[0] = 1;
for (auto coin : coins) {
for (int j = coin; j <= amount; ++j) {
dp[j] += dp[j - coin];
}
}
return dp[amount];
}
};

• class Solution:
def change(self, amount: int, coins: List[int]) -> int:
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
for j in range(coin, amount + 1):
dp[j] += dp[j - coin]
return dp[-1]


• func change(amount int, coins []int) int {
dp := make([]int, amount+1)
dp[0] = 1
for _, coin := range coins {
for j := coin; j <= amount; j++ {
dp[j] += dp[j-coin]
}
}
return dp[amount]
}

• function change(amount: number, coins: number[]): number {
let dp = new Array(amount + 1).fill(0);
dp[0] = 1;
for (let coin of coins) {
for (let i = coin; i <= amount; ++i) {
dp[i] += dp[i - coin];
}
}
return dp.pop();
}