P

```
/**
* @params {Array} coins
* @params {Number} amount
*/
export const change = (coins, amount) => {
// Create and initialize the storage
const combinations = new Array(amount + 1).fill(0)
combinations[0] = 1
// Determine the direction of smallest sub-problem
for (let i = 0; i < coins.length; i++) {
// Travel and fill the combinations array
for (let j = coins[i]; j < combinations.length; j++) {
combinations[j] += combinations[j - coins[i]]
}
}
return combinations[amount]
}
/**
* @params {Array} coins
* @params {Number} amount
*/
export const coinChangeMin = (coins, amount) => {
const map = { 0: 1 }
for (let i = 1; i <= amount; i++) {
let min = Infinity
for (const coin of coins) {
if (i < coin) continue
min = Math.min(min, 1 + map[i - coin])
}
map[i] = min
}
return map[amount] === Infinity ? -1 : map[amount] - 1
}
```

Given a value `N`

, if we want to make change for `N`

cents, and we have infinite supply of each of `S = {S1, S2, .. , Sm}`

valued coins, how many ways can we make the change? The order of coins doesn’t matter.

Let the `dp[i]`

be the length of the coin change of prefix `N[1..i]`

. Our answer is `dp[N]`

.
We fill `dp[0]`

as 1 because there is only one way to get 0 coins (We pick no coins).

Now let's try calculate `dp[i]`

for any `i`

. `dp[0..i]`

will store each sub problems from `0`

to `N`

. That's why the answer will be `dp[N]`

. First, we need to iterate each coin types to pick them one-by-one. Then we iterate the sub problems from current coin that we pick before to `N`

cents. That's why we must make dp table with `N`

columns.

This is the codes for the Coin Change algorithm:

```
for coin_val in S:
for i in range(coin_val, n + 1):
dp[i] += dp[i - coin_val]
```

In the second iteration, for every cent that can be exchanged, we take it by subtracting the i-th column by the value of the coin we take and adding it into the current column. So `dp[i]`

will store the current sub problem.

`O(N * S)`

in any case

`O(N)`

- simple implementation. We only need 1D array to store the answer.

Let's say we have 3 coin types `[1,2,3]`

and we want to change for `7`

cents. So we will define our table like this.

```
[1, 0, 0, 0, 0, 0, 0, 0]
```

0th column will store 1 since there is only one way to get 0 cents.

- For the first iteration we take a coin that has a value of 1. Then for all sub problems, there is only one way to make change. For 7 cents, the only way is
`{1,1,1,1,1,1,1}`

. On the final iteration, our table be like:

```
[1, 1, 1, 1, 1, 1, 1, 1]
```

- For the second iteration, we take a coin that has a value of 2. From here, all sub problems that can be divided by 2 will store another new way to make change. So, when the iteration stopped at 2nd column it will be like
`dp[2] += dp[0]`

. We know that`dp[0]`

stored a value of 1. Thus, dp[2] will store the value of`1 + 1 = 2`

. From here we know that for 2 cents, there are 2 ways`{1,1}`

and`{2}`

. And this operation will continue. Now our table be like:

```
[1, 1, 2, 2, 3, 3, 4, 4]
```

4 ways to make 7 cents using value of 1 and 2. `{{1,1,1,1,1,1,1}, {1,1,1,1,1,2}, {1,1,1,2,2}, {1,2,2,2}}`

- For the final iteration (3rd iteration), we take a coin that has a value of 3. Like before, now all the columns that can be divided by 3 will store another new way. And the final result will be like:

```
[1, 1, 2, 3, 4, 5, 7, 8]
```

So the final answer is **8**. 8 ways to make change of 7 cents using all coin types. `{{1,1,1,1,1,1,1}, {1,1,1,1,1,2}, {1,1,1,2,2}, {1,2,2,2}, {1,1,1,1,3}, {1,3,3}, {2,2,3}, {1,1,2,3}}`